SSC MATHEMATICS: Chapter 6 )

Example 1) For the AP : (1/4),(-1/4),(-3/4),(-5/4)......., write the first term "a" and the common difference"d". And find the 7th term.

sol) Here, a = (1/4)

d = ( -1 ) - ( 1 )
    4    4

d = -1
    2

Remember that we can find "d" using any two consecutive terms, once we know that the numbers are in AP.

The 7th term would be:-

Since the 7th term is 3 steps away from the 4th term(-5/4), Add "3 times" the common difference to get the value of 7th term

-5 -1 -1 -1 = -11
4 2 2 2 4

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Example 2) Which of the following forms an AP? If they form AP then write next two terms?

(!) 4,10,16,22

Sol) We have

a2-a1 = 10-4 =6

a3 - a2 = 16-10 = 6

a4 - a3 = 22-16 = 6

i.e, ak+1 - ak is same every time.

So, the given list of numbers forms an AP with the common difference d=6.

The next two terms are:

22+6 =28

28 +6 = 34

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(!!) a2 - a1 = -1 -1 = -2

a3 - a2 = -3 -(-1) = -3+1 =-2

a4 - a3 = -5 - (-3) = -5+3 = -2

i.e, ak+1 - ak is same every time.

So, the given list of numbers form an AP with the common difference d=-2

The next two terms are:-

-5 +(-2) = -7

-7 +(-2) = -9

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(!!!) d1 =a2 - a1 = 2-(-2) = 4

    d2= a3 - a2 = -2 -2 =-4

d1 =/= d2, the given list of numbers do not form an AP.

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(IV) d1 = a2-a1 = 1-1=0

d2 = a3-a2 = 1-1=0

d3 = a4 -a3 = 2-1 =1

Here, d1 = d2 =/= d3

So, the given list of numbers do not form an AP.

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(V) d1 = a2 - a1 = 2x -x = x

d2 = a3 - a2 = 3x-2x = x

d3 = a4 - a3 = 4x - 3x = x

i.e, ak+1 - ak is same every time.

.^. So, the given list form an AP.

The next two terms are:-

4x + x = 5x

5x + x = 6x

Note:- To find "d" in the AP: 6,3,0,-3....we have subtracted 6 from 3 and not 3 from 6. We have to subtract the kth term from the (k+1)th term even if the (k+1)th term is smaller and to find "d" in a given A.P. We need not find all of
a2-a1 , a1-a2......its enough to find only one of them.

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Exercise - 6.1

1) in which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(1) The taxi fare after each km when the fare is 20Rs for the first km and rises by 8Rs for each additional km.

sol) It is an Arithmetic progression or AP

reason :- Taxi for 1st km = 20rs

=> for 2km = 20 + 8 = 28

=> for 3km = 28 + 8 = 36

.^. , series is 20, 28, 36,,,,,,,,

difference between 2nd term and 1st term

=> 28 - 20

=> 8

Difference between 3rd and 2nd term

=> 36 - 28 = 8

Since difference is same, It is an AP

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2) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

sol) Let volume of cylinder be 16 liters

If we remove (1 / 4) air from out of 16 liters .how much air will remain?

i.e = 1 / 4 * 16 = 4 liters (removed)

Air present after first removal = 16-4 = 12 liters(after removal)

Again

Air removed 1 / 4 * 12 = 3 liters (removed)

Air present after 2nd removal = 12 - 3 = 9 liters

The amount of air present in cylinder in the series

16, 12, 9.......

difference between 3nd and 2nd air removal

=> 9 - 12

=> - 3

Difference between 2nd and 1st air removal

=> 12 - 16

=> -4

Since difference is not same -3, and - 4

.^. it is not an Arithmetic Progression

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3) The cost of digging a well after every meter of digging , when it costs Rs. 150 for the first meter and rises by 50 for each subsequent meter.

sol) Cost of digging first meter = 150

=> for every meter 50Rs added :-

Cost of digging 2 meter = 150 + 50 = 200

Cost of digging 3 meter = 200 + 50 = 250

So, the series is

150, 200, 250,,300.........

Difference between third and second meter = 250 - 200 = 50

Difference between 2nd and 1st meter = 200 - 150 = 50

Since difference is same, It is an Arithmetic progression.

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4) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum

sol) Given :- Initial deposite = 10,000Rs

Calculating :- how much amount increased per year after adding 8% interest

Interest earned in first year = 10,000 * 8%

=> 10,000 * 8 / ~~100~~

=> 8*100 = 800

Total amount after 1 year : 10,000 + 800 = 10800 Rs

(*) Interest earned in 2nd year = 10800 * 8 / ~~100~~

=> 108 * 8 = 864

Total amount after 2nd year = 10800 + 864

=> 11,664

(*)Interest earned in 3rd year = 11,664 * 8 / 100

=> 933.12

Total amount after 3rs year = 11,664 + 933.12

=> 12597.12

We got the amount every year in s series at 8% is :-

10800, 11664, 12597.12

Finding difference:-

3rd and 2nd year = 12697.12 - 11664 = 1033.12

2nd and 1st year = 11664 - 10800 =864

Since difference is not same

864 =/= 1033.12

.^. it is not in Arithmetic progression
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2) Write first four terms of the AP, when the first term "a" and common difference "d" are given as follows:-

(1) a = 10, d = 10

sol) Given :- 1st number = 10 and difference between 1st and 2nd number = 10

now,

2nd term : 10 + 10 = 20

3rd term = 20 + 10 = 30

4th term = 30 + 10 = 40

.^. , first four term of AP = 10, 20, 30, 40

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(2) a = -2, d = 0

sol) given :- 1st number = -2 and difference between 1st and 2nd number = 0

now,

2nd term = -2 + 0 = -2

3rd term = -2 + 0 = -2

4th term = -2 + 0 = -2

.^. , first four term of AP = -2, -2, -2, -2

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3) a = 4, d = -3

sol) Given :- 1st number = 4 difference between 1st and 2nd number = -3

2nd term = 4 - 3 = 1

3rd term = 1 -3 = -2

4th term = -2 -3 = -5

.^. First 4 term of AP = 4 , 1, -2 , -5

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4) a = -1, d = 1/2

sol) Given :- 1st number = -1 difference b/w 1st and 2nd no = 1/2

2nd term = -1 + 1 /2 => -2 + 1 => -1 /2
2

3rd term = -1/2 + 1/2 = 0

4th term = 0 + 1/2 = 1/2

.^. , the first four terms of AP are = -1, -1/2, 0, 1/2

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5) a = -1.25, d = -0.25

sol) first term = -1.25

2nd term = -1.25 + ( - 0.25) = -1.25 - 0 . 25 => -1.50

3rd term = -1.50 + (-0.25) => -1.50 - 0.25 => -1. 75

4th term = -1.75 + ( - 0.25) => -1.75 - 0.25 =>-2.00

.^. First four term of AP = -1.25, -1.50, -1.75, -2.00

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(3) For the following AP's write the first term and the common difference:

(1) 3,1, -1, -3...

sol) first term(a) = 3

common difference(d) = 2nd term - 1st term

=> 1 - 3

=> -2

Hence, a = 3, d = -2

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(2) -5, -1, 3, 7......

sol) First term(a) = -5

=> common difference(d) = 2nd term - 1st term

=> -1 - (-5)

=> -1 + 5

=> 4

Hence a = -5, d = 4

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(3) 1/3 , 5/3, 9/3, 13/3

sol) First term (a) = 1/3

Common difference(d) = 2nd term - 1st term

=> 5 /3 - 1/3

=> 5 - 1
3

=> 4 / 3

Hence a = 1/3, d = 4 /3

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(4) 0.6, 1.7, 2.8, 3.9....

sol) First term(a) = 0.6

common difference (d) = 2nd term - 1st term

=> 1.7 - 0.6

=>1.1

Hence a = 0.6, d = 1.1

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(4) Which of the following are AP's? If they form an AP, find the common difference "d" and write three more terms.

(1) 2,4,8,16.....

sol)  Finding If they are AP or not :-

Difference between 2nd and 1st term = 4 - 2 = 2

Difference between 3rd and 2nd term = 8 - 4 = 4

Since the difference is not same 2 =/= 4

Hence, it is not an AP.

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(2) 2, 5/2, 3, 7/2

sol) Finding if the numbers are AP or not :-

Difference b/w 2nd and 1st term = 5/2 - 2

=> 5 - 4
2

=> 1 / 2

Difference b/w 3rd and 2nd term = 3 - 5 /2

=> 6 - 5
2

=> 1 / 2

Since difference is same 1/2 = 1/2. It is an AP

common difference (d) = 1/2

Finding next 3 more terms:-

Finding 5th term => 4th term + d

=> 7 /2 + 1/2

=> 7 + 1
2

=> 8 /2

=> 4

Finding 6th term => 5th term + d

=> 4 + 1/2

=> 8 + 1
2

=> 9 /2

Finding 7th term => 6th term + d

=> 9 /2 + 1 /2

=> 10 / 2

=> 5

Hence next 3 more terms are = 4, 9/2, 5

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(3) -1.2, -3.2, -5.2, -7.2

sol) Finding if the series are in AP or not :-

Common difference (d1) = 2nd term - 1st term

=> -3.2 - ( -1.2)

=> - 3.2 + 1.2

=> -2.0

Common difference(d2) = 3rd term - 2nd term

=> -5.2 - (-3.2)

=> -5.2 + 3.2

=> -2.0

Since difference is same d1 = d2

Hence, its an AP

Finding next 3 more terms :-

Finding 5th term => 4th term + d

=> - 7.2 + (-2)

=> -9.2

6th term :- 5th term + d

=> -9.2 + (-2)

=> -11.2

7th term :- 6th term + d

=> -11.2 + (-2)

=> -13.2

Hence 4th, 6th and 7th terms are = -9.2, -11.2, -13.2

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(4) -10, -6, -2, 2...

sol) Finding if the series are in AP or not :-

common difference (d1) = 2nd term - 1st term

=> -6 - ( -10)

=> -6 + 10

=> 4

d2 = 3rd term - 2nd term

=> -2 - ( -6)

=> -2 + 6

=> 4

Since d1 = d2

Series are in AP

finding next 3 more terms

5th term => 4th term + d

=> 2 + 4 = 6

6th term => 5th term + d

=> 6 + 4 = 10

7th term => 6th term + d

=> 10 + 4 = 14

Hence 5th, 6th and 7th terms are = 6, 10 , 14

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(5) 3, 3+ _/2, 3+ 2_/2, 3+3_/2

sol) Finding if the series are in AP or not

d1= 2nd term - 1st term

=> 3 + _/2 - 3

=> _/2

d2 = 3term - 2nd term

=> (3+2_/2) - (3+ _/2)

=>3+2_/2 -3-_/2

=> 2_/2 - _/2

=> _/2 (2 -1)

=> _/2(1) => _/2

Since d1 = d2

Series are in AP

finding next 3 more terms

5th term => 4th term + d

=> (3 + 3_/2) + _/2

=> 3 + _/2(3 + 1)

=> 3 + -/2(4)

6th term => 5th term + d

=> 3 + 4_/2 + _/2

=> 3 + _/2(4 + 1)

=> 3 + _/2(5)

= 3 + 5_/2

7th term = > 6th term + d

=> 3 + 5_/2 + _/2

=> 3 + _/2 (5 + 1)

=> 3 + 6_/2

hence 5th ,, 6th and 7th term = 3 + -/2(4), 3+5_/2, 3+6_/2

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6) 0.2, 0.22, 0.222, 0.2222....

sol) Finding if the series are in AP or not;-

d1 => 2nd term - 1st term

=> 0.22 - 0.20

=> 0.02

d2=> 3rd term - 2nd term

=> 0.222 - 0.22

=> 0.002

So, difference is not same d1 =/= d2

Hence this is not a AP

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7) 0, -4, -8, - 12

sol) Finding if the series are in AP or not :-

d1 => 2nd term - 1st term

=> (-4) - 0

=> -4

d2 => 3rd term - 2nd term

=> -8 - (-4)

=> -8 + 4

=> -4

Since difference is same , d1 == d2

.^. it is an AP

finding 5th, 6th, & 7th:-

5th term = 4th term + d

=> -12 +(-4)

=> -12 - 4

=> - 16

6th term = 5th term + d

=> -16 + (-4)

=> -20

7th term = 6th term + d

=> -20 + (-4)

=> -20 - 4

=> - 24

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8) -1/2, -1/2, -1/2, -1/2....

sol) Finding if the series are in AP or not :-

d1 => 2nd term - 1st term

=> (-1/2) - (-1/2)

=> (-1/2) + 1/2

= 0

d2 = > 3rd term - 2nd term

=> (-1/2) -(-1/2)

=> -1/2 + 1/2

=> 0

Since difference is same, d1 = d2

.^. it is an AP

finding next 3 more terms ;-

5th term => 4th term + d

=> -1/2 + 0

=> -1/2

6th term => 5th term + d

=> -1/2 + 0

=> -1/2

7th term => 6th term + d

=> -1/2 + 0

=> -1/2

.^. , 5th, 6th & 7th terms = -1/2 , -1/2, -1/2

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(9) 1, 3, 9, 27

sol) Finding if the series are in AP or not :-

d1 = 2nd term - 1st term

=> 3 - 1

=> 2

d2 = 3rd term - 2nd term

=> 9 - 3

=> 6

Since, difference is not same d1 =/= d2

Hence it is not an AP

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10) a, 2a, 3a, 4a

sol) Finding if the series are in AP or not :-

d1 = 2nd term - 1st term

=> 2a - a

=> a

d2 = 3rd term - 2nd term

=> 3a - 2a

=> a

Since difference is same d1 = d1

.^. it is an AP

Finding next 3 more terms

5th term :- 4th term + d

=> 4a + a

=> 5a

6th term :- 5th term + d

=> 5a + a

=> 6a

7th :- 6th term + d

=> 6a + a

=> 7a

Hence 5th, 6th, & 7th terms = 5a, 6a, 7a

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11) a, a^2, a^3, a^4....

sol) Finding if this series are in AP or not :-

d1 = 2nd term - 1st term

d1 = a^2 - a

d1 = a(a - 1)

d2= 3rd term - 2nd term

d2 = a^3 - a^2

d2 = a^2(a -1)

Since d1 =/= d2 ..It is not an AP

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12) _/2, _/8, _/18, _/32....

sol) Finding if they are in AP or not.

d1 = 2nd term - 1 term

=> _/8 - _/2

=> _/(2 * 2* 2) - _/2

=> _/2 * (2)^2 - _/2

=> 2_/2 - _/2

=> _/2(2 - 1)

=> _/2(1)

=> _/2

d2 = 3term - 2nd term

=> _/18 - _/8

=> _/3 * 3 * 2 - _/ 2 * 2* 2

=> 3_/2 - 2_/2

=> _/2 (3 -2)

=> _/2 (1)

=> _/2

Since d1 == d2, It is an AP

Finding 5th, 6th and 7th term

5th term = 4th term + d

=> _/32 + _/2

=> _/4 * 4 * 2 + _/2

=> _/(4)^2 * 2 + _/2

=> 4_/2 + _/2

=> _/2(4 + 1)

=> _/2(5)

=> 5_/2

6th term = 5th term + d

=> 5_/2 + _/2

=> _/2( 5 + 1)

=> 6_/2

7th term = 6th term + d

=> 6_/2 + _/2

=> 7_/2

Hence the next 3 terms are = 5_/2, 6_/2, 7_/2

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13) _/3, _/6, _/9, _/12....

sol) Finding if the series are in AP or not

D1 = 2nd term - 1st term

=> _/6 - _/3

=> _/(2 *3) - _/3

=> _3 (_/2 - 1)

d2 = 3rd term - 2nd term

=> _/9 - _/6

=> _/3 * 3 - _/ 2*3

=> _/3 ( _/3 - _/2)

The difference is not same d1 =/= d2

Hence , it is not an AP.

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Exercise - 6.2

1) Fill in the blanks in the following table, given that "a" is the first term, "d" the common difference and "an" the nth term of the AP

(1) a = 7, d = 3, n = 8, an=?

sol)  We have

Formula :- an = a + (n - 1) d

=> an = 7 + (8 -1) *3

=> an = 7 + 7*3

=> an = 7 + 21

.^. an = 28

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(2) a = -18, d = ?, n = 10, an = 0

sol) We have

Formula :- an = a + (n - 1) d

=> 0 = -18 + ( 10 -1) * d

=> 0 = -18 + (9) * d

=> 0 + 18 = 9d

=> 18 = d
9

=> 2 = d

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(3) a=?, d= -3, n = 18,  an = -5

sol) We have

Formula :- an = a + (n - 1) d

=> -5 = a + (18 -1) * -3

=> -5 = a + (17) * (-3)

=> -5 = a + (-51)

=> -5 = a -51

=> -5 + 51 = a

=> 46 = a

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(4) a = - 18.9, d = 2.5, n =?, an = 3.6

sol) We have

Formula :- an = a + (n - 1) d

=.3.6 = -18.9 + ( n -1) * 2.5

=> 3.6 = -18.9 + 2.5n - 2.5

=>3.6 = -21.4 + 2.5n

=> 3.6 + 21.4 = 2.5n

=> 25 = 2.5n

=> 25 = n
2.5

=> 10 = n
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(5) a = 3.5, d = 0, n = 105,  an =?

sol ) We have

Formula :- an = a + (n - 1) d

=> an = 3.5 + ( 105 - 1) * 0

=> an = 3.5 + 0

Hence an = 3.5

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Example 3: Find the 10th term of the AP : 5,1,-3,-7....

Sol) Here a = 5, d=1-5 =-4

and n= 10

We have an = a + (n-1)d

So, a10 = 5 + (10-1)(-4)

=> 5 + (9)(-4)

=> 5 -36

.^. -31 is the 10th term of the given AP.

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Example 4) Which term of the AP : 21,18,15....is -81? Is there any term "0"? Give reason for your answer.

sol) Here a = 21

d = 18-21 = -3

an = -81, we have to find "n".

As an = a +(n-1)d,

we have

-81 = 21 +(n-1)(-3)

-81 = 21 -3n +3

-81 = 24 - 3n

3n = 81 +24

n = 105
3

n = 35

.^., the 35th term of the given AP = -81

Next, we want to know if there is any "n" for which an = 0. If such an "n" is there, then

21 + (n-1)(-3) = 0

3(n-1) = 21

n-1 = 7

n = 7+1 = 8

So, the 8th term is "0"

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Example 5) determine the AP whose 3rd term is 5 and 7th term is 9.

sol) We have

a = a + (3-1)d => a +2d = 5 ----(1)

a = a +(7-1)d => a + 6d = 9 ---(2)

Solving the pair of linear equations (1) and (2), we get

a = 3, d = 1

Hence, the required AP is 3,4,5,6,7........

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Example 6): Check whether 301 is a term of the list of numbers 5,11,17,23.....

Sol) We have :

d1 = a2 - a1 = 11-5 = 6

d2 = a3 - a2 = 17-11=6

d3 = a4 - a3 = 23 -17 =6

As(ak+1 - ak) is the same for k = 1,2,3,etc., the given list of numbers is an AP.

Now, for this AP we have a=5 and d=6.

We choose to begin with the assumption that 301 is a term, say, the nth term of this AP.

We will see if an "n" exists for while an = 301.

We know

an = a + (n-1)d

So, for 301 to be a term we must have

301 = 5 + (n+1) * 6

or 301 = 6n -1

so, n = 302
6

=>n = 151
3

But"n" should be a positive integer(why?)

So, 301 is not a term of the given list of numbers.

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Example 7): How many two-digit numbers are divisible by '3"?

sol)The list of 2 digit numbers divisible by "3" is :

12, 15, 18,......99

Is this an AP? Yes it is

Here,

a = 12

d = 15-12 = 3

an = 99

As an = a +(n-1)d

We have

99 = 12 + (n-1) *3

=> 87 = (n-1) *3

=>n-1 = 87
   3

=> n-1 = 29

=> n = 30

So, there are 30 two-digit numbers divisible by "3"

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Example 8) Find the 11th term from the last of the AP series given below:

AP : 10,7,4..., -62.

sol) Here, a =10

d = 7-10 = -3

L = -62.

Where, L = a +(n-1)d

To find the 11th term from the last term, we will find the total number of terms in the AP.

So,

-62 = 10 +(n-1)(-3)

-72 = (n-1)(-3)

n-1 = 24

n = 25

So, there are 25 terms in the given AP.

The 11th term from the last will be the 15th term of the series.(Note that it will not be the 14th term. why?)

So,

a15 = 10 + (15-1)(-3)

=> 10 -42

=> -32.

i.e, the 11th term from the end = -32

Note :- The 11th term from the last is equal to 11th term of the AP with first term"-62" and the common difference "3"

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Example 9) A sum of 1000rs is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years

sol) We know that the formula to calculate simple interest is given by

Simple Interest = P * R * T
                                    100

So, the interest at the end of 1st year:-

=> 1000 * 8 * 1 => 80
             100

Similarly, we can obtain the interest at the end of the 2nd, 3rd,4th and so on.

So, the interest(in Rs) at the end of the

1st, 2nd, 3rd,....years respectively are

80, 160, 240.........

It is an AP as the common difference in the list = 80.

i.e,   d =80, Also a= 80

So, to find the interest at the end of 30 years, we shall find a30.

Now,

a30 = a + (30-1)d

=> 80 + (29)*80

=> 2400

So, the interest at the end of 30 years will be= 2400 rs

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Example 10) In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on.There are 5 rose plants in the row. How many rows are there in the flower bed?

sol) The number of rose plants in the 1st, 2nd, 3rd....rows are :

23, 21,19......5

It forms an AP(why?)

let the number of rows in the flower bed be "n"

Then,

a = 23

d = 21-23 = -2

an = 5

We have

an = a + (n-1)d

5 = 23 + (n-1)(-2)

=> -18 = -2n +2

=> -20 = -2n

n = 10

So, there are 10 rows in the flower bed.

(2) Find the

(1) 30th term of the A.P 10, 7, 4,.......

sol) Given series = 10, 7, 4......

Finding 30th term :-

a = 10, d = 7 - 10 = -3 , n = 30 , an =?

We have
Formula : an = a+ (n-1) d

=> a30 = 10 + (30-1) * (-3)

=>a30 = 10 + (29) * (-3)

=> a30   = 10 +( -87)

=> a30   = 10 - 87

=> a30 = -77

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(2) 11th term of the A.P : -3, -1/2, 2.....

sol) Given series = -3, -1/2, 2.....

Finding 11th term :-

a = -3,

d = -1/2 -(-3)

=> -1/2 + 3 => -1 + 6 = 5 /2
2
n = 11

a11 = ?

We have
Formula : an = a+ (n-1) d

a11 = -3 + (11-1) * 5 /2

a11 = -3 + (~~10)~~ * 5
2

a11 = -3 + (5) * 5

a11 = -3 + 25 => 22

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(3) Find the respective terms for the following APs.

(1) a1 = 2, a3 = 26 find a2

sol) let a2 = x

then the series will be = 2 , x , 26....

Given :- terms are in APs

.^. common difference will be same

d1 = d2

a2 - a1 = a3 - a2

x - 2 = 26 - x

x + x = 26 +2

2x = 28

x = 14

.^. The series is 2, 14, 26

******************************************************

(2) a1=?, a2 = 13, a3 = ?, a4 = 3

sol) Since its in Ap...common difference will be same

finding a3:-

a4 - a3 = a3 - a2

3 - a3 = a3 - 13

3 + 13 = a3 + a3

16 = 2 a3

16/ 2 = a3

8 = a3

finding a1:-

a3 - a2 = a2 - a1

8 - 13 = 13 - a1

-5 = 13 - a1

-5 -13 = -a1

-18 = a1

a1 = 18

.^. The given series = 18, 13, 8, 3

*******************************************************
(3) 5, a2, a3, 9 1/2

sol) Given a = 5

a4 = 9 * 1/2 => 2(9) + 1 = 19
2 2

We have => an = a + (n -1) d

first finding common difference

a4 = 19/2, a = 5, n = 4, d =?

a4 = a + (n -1) d

19 / 2 = 5 + (4 -1) * d

19/ 2 = 5 + (3) d

19 - 5 = 3d
2

19 - 10 = 3d
2

9 = 3 d
2

3 = d
2

Hence , a = 5, d = 3 /2

finding a2

a2 = a + (n -1) d

=> 5 + (2-1) * 3/2

=> 5 + (1) * 3/2

=> 5 + 3/2

=> 10 + 3
2

a2 = 13  = 6 1
2 2

finding a3

a3 = a + (n -1) d

=> 5 + (3-1) * 3/2

=> 5 + (2) * 3/2

=> 5 + 3

a3 = 8

Hence, the series is = 5, 6 1,   8 , 9 1
2 2

************************************************************

(4) a1 = -4, a6 = 6 find a2, a3, a4, a5

sol) Step 1 :- First finding common difference

Given :- a = -4, a6 = 6

We have formula :- an = a + (n-1) d

a6 = -4 + (6-1) d

6 + 4 = (5) d

10 =5d

2 = d

Now, we got a = -4, d = 2

we can find other series

a2 = a + (n -1) d

=> -4 + ( 2-1) d

=> -4 + (1) d

=> -4 + (1) 2

=> -2

Finding a3

a = -4, d = 2, n = 3, a3=?

a3 = a + (n-1) d

=> -4 + ( 3-1)*2

=> -4 + (2) * 2

= -4 + 4

a3 = 0

Finding a4

a= -4 ,, d = 2, n= 4, a4=?

a4 = a + (n -1)d

=> -4 + (4-1) *2

=> -4 + (3) *2

=> -4 + 6

a4 = 2

Finding a5

a = -4, d = 2, n = 5, a5 = ?

a5 = a + (n-1) d

=> -4 + (5-1) *2

=> -4 + (4) *2

=> -4 + 8

a5 = 4

Hence the series is :- -4, -2, 0, 2, 4, 6

*********************************************************

(5) a2 = 38, a6 = -22 find a1, a3, a4, a5

sol) Step 1:- First finding common difference

a2 = 38, a6 = -22

We have formula :- an = a + (n-1) d

a2 = a + (2-1) d

38 = a + (1) d

a = 38 -d--------(1)

a6 = a + (6-1) d

-22 = a + ( 5) d

-22 = a + 5d

-22 -5d = a -------- (2)

from (1) and (2)

38 - d = -22 - 5d

38 + 22 = d - 5d

60 = -4d

-15 = d

putting "d" in (1) we get "a" value

a = 38 - d

=> 38 - (-15)

=> 38 + 15

=> 53

Hence we got a = 53 , d= -15

We need to find a3, a4, a5

a3 = a + (3-1) *(-15)

=> 53 + (2) * (-15)

=> 53 - 30

=> 23

a4 = a + (4-1)d

=> 53 + (3) (-15)

=> 53 - 45

=> 8

a5 = a + (5-1) d

=> 53 + (4) (-15)

=> 53 - 60

= - 7

Hence , the series is = 53, 38, 23, 8, -7, -22

******************************************************

4) Which term of the AP : 3, 8, 13, 18,...... is 78?

sol) Here a =3,

common difference (d) = 2nd term - 1st term

=> 8 - 3 => 5

Hence an= 78, a = 3, d = 5

Putting these in formula

an = a + (n -1) d

78 = 3 + ( n-1) *5

78 - 3 = (n-1) *5

75 = (n-1) *5

75 = n -1
5

15 = n -1

15 + 1 =n

16 = n

.^. the 16th term of AP is 78

********************************************************

5) Find the number of terms in each of the following APs:

(1) 7, 13, 19,......205

sol) Given:- a = 7 ,

common difference (d) = 13 - 7 = 6

an = 205

We need to find "n":-

an = a + (n-1) d

205 = 7 + ( n-1) *6

205 - 7 = (n-1) *6

198 = (n-1) *6

198 = n-1
6

33 = n-1

33 + 1 = n

34 = n

Hence there are 34 terms in the given AP

**************************************************************

(2) 18, 15 1, 13,.......................- - 47
2

sol) an = - 47

a = 18

d = 15 1 = 31   - 18
2 2

31 - 36 = -5
   2 2

We need to find "n"

an = a + (n-1) d

-47 = 18 + (n-1) * ( -5/2)

-47 - 18 = (n-1) * (-5/2)

-65 = (n-1) (-5/2)

-65 * 2= (n-1)
-5

26 = n-1

n = 26 + 1

n = 27

Hence the number of terms in the given AP is 27

*********************************************************

6) Check whether , -150 is term of the AP : 11, 8, 5, 2...

sol) Lets assume it is nth term of AP

so, an = - 150

Also a = 11

d = 8 -11 = -3

Now using formula

an= a + (n-1) d

-150 = 11 + (n-1) * (-3)

-150 = 11 + (n-1) *(-3)

-150 - 11 = (n-1) (-3)

- 161 = (n-1) (-3)

-161 = -3n + 3

-161 - 3 = -3n

-164 = -3n

164 = n
3

54.33 = n

n = 54.33

Since "n"is decimal

Hence, -150 is not term of AP

*************************************************************

7) Find the 31th term of an AP whose 11th term is 38 and the 16th term is 73.

sol) Given :- 11th term is 3873

=> a11 = a + ( 11 -1) d

=> 38 = a + (10) d

=> 38 - 10d = a -------(1)

Given :- 16th term is 73

=> a16 = a + (16-1) d

=> 73 = a + (15) d

=> 73 - 15d = a ---------(2)

equation (1) and (2)

38 -10d = 73 - 15d

38 -73 = -15d - 10d

-35 = -5d

7 = d

Putting the value of "d" in (1)

a = 38 - 10d

a = 38 - 10 *7

a = 38 - 70

a = -32

We need to find the 31th term

So, n =31, a = - 32, d = 7

We need to find "an"

an = a + (n-1) d

a31 = -32 + ( 31-1) *7

=> -32 + 30*7

=> -32 + 210

=> 178

.^. , the 31th term of AP is 178

************************************************************

8) if the 3rd and 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?

sol) We know that

If the 3rd term is 4;-

a3 = 4 , n =3

an =  a + (n-1) d

a3 = a + (3-1)*d

4 = a + (2) d

4 - 2d = a---------(1)

If the 9th term is -8

a9 = -8, n = 9

an =  a + (n-1) d

a9 = a + ( 9-1)d

-8 = a + 8d

-8 - 8d = a ------- (2)

equating (1) and (2)

4 - 2d = -8d - 8

-2d + 8d = -8 -4

6d = -12

d = -12 /6

d = -2

substitute "d" value in (1)

a = 4 -2d

a = 4 - 2(-2)

a = 4 +4

a = 8

Lets assume its nth term in the series which is "zero"

i.e, an = 0,

a = 8, d = -2,

Using the formula we can get "an " value

an = a + (n-1) d

0 = 8 + ( n-1)* -2

0 = 8 + (n*-2) + 2

0 = 8 -2n + 2

10 - 2n = 0

10 = n

n = 5

.^. 5th term of AP is "0"
***********************************************************

(9) The 17th term of an AP exceeds its 10th term by 7. Find the common difference .

sol) We have

an = a + (n-1) d

17th term:-

a17 = a + (17-1) d

a17 = a + 16d ------- (1)

10th term :-

a10 = a + (10-1) d

a10 = a + 9d -----(2)

condition :- 17th term exceeds its 10th term by 7

i.e a17 = a10 + 7

=> a17 - a10 = 7-----(3)

Now substitute a17 & a10 value in (3)

=>( a + 16d) - ( a + 9d) = 7

=> a - a + 16d - 9d = 7

=> 0+ 7d = 7

=> 7d = 7

=> d = 1

.^. , common difference of the given AP is "1"

******************************************************

10) Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms.

sol) We know that :- an = a + (n-1) d

Given:- Two APs have the same common difference

So, 100th term will be

a100 = a + ( 100-1) d

a100 = a + 99d ------>( 100th term of 1st AP)

Let 2nd AP be denoted as

b100 = b + (100-1)d

b100 = b + 99d -------> ( 100th term of 2nd AP)

condition:-common difference(d) is same(100) in two AP

a100 - b100 = 100

(a + 99d) - ( b +99d) = 100

a + 99d - b - 99d = 100

a - b+ 99d - 99d = 100

a - b = 100 ----- (1)

Finding difference between 1000th terms

an = a + (n-1) d

=> a1000 - b1000

=> [ a + (1000-1)d ] - [ b + (1000-1)d]

=> [ a + 999d] - [ b + 999d]

=> a + 999d - b - 999d

=> a- b ----(2)

.^. the difference between 1000th terms is 100

**********************************************************

11) How many three-digit numbers are divisible by 7 ?

sol) first lower 3 digit number start from 100

(*) is 100 divisible by "7"

=> 100 = 1.57
7

(*) is 101 divisible by "7"

=> 101 = 14.42
7

(*) is 102 divisible by "7"

=> 102 = 14.57
7

(*) is 103 divisble by "7"

=> 103 = 14.71
7

(*) is 104 divisible by "7"

=> 104 = 14.85
7

(*) is 105 divisible by "7"

=> 105 = 15
7

"105" is the first three digit number divisible by "7"

Now finding the last three digit number divisible by "7"

is 999 divisible by "7"

=> 999 = 142.714
7

(*) is 998 divisible by "7"

=> 998 = 142. 57
7

(*) is 997 divisible by "7"

=> 997 = 142.42
7

(*) is 996 divisible by "7"

=> 996 = 142.28
7

(*) is 995 divisible by "7"

=> 995 = 142.14
7

(*) is 994 divisible by "7"

=> 994 = 142
7

"994" is the last three-digit number divisible by "7"

As we can clearly see three digit number is divisible "7" after every 7th number

i.e, 100, 101,102,103,104,105, 106, 107,108,109,110,111,112......

So, series will be

105, 112, 119,........994

Since the difference is same(7), it is an AP

We need to find number of terms ,"n"?

here an = 994,

=> a = 105

=> d = 7

=> n = ?

Putting values in formula

an = a + (n-1) d

994 = 105 + (n - 1) 7

994 = 105 + 7n - 7

994 - 105 + 7 = 7n

896 = 7n

896 = n

128 = n

.^. total 128 three digit numbers are divisible by "7" ina series

*************************************************************

12) How many multiples of "4" lie between 10 and 250

sol) Multiple of "4' : 4 , 8, 12.....

Finding first multiple of "4" start from "10"

(*) 10 = 2.5
4

(*) 11 = 2.75
4

(*) 12 = 3
4

We got first multiple of "4" i.e, "12"

Finding last multiple of "4" which end under "250"

(*) 250 = 62.5
4

(*) 249 = 62.25
4

(*) 248 = 62
4

We got the last multiple of "4" i.e, 248

We can observe after every three number fourth digit is a multiple of "4"

i.e, 10, 11, 12, 13,14,15, 16.........

So the series is

12, 16, 20......248

Since difference is same, it is an AP

Here a = 12, d = 4, last term(an) = 248, n =?

We have

an =  a + (n-1) d

248 = 12 + (n-1)4

248 = 12 + 4n -4

248 -12+4 = 4n

240 = 4n

240 = n
4

n = 60

So, 60 multiples of "4" lie between 10 and 250

********************************************************

13) For what value of "n", are the nth terms of two APs : 63, 65, 67,...
and 3, 10, 17.... equal?

sol) 1st AP :- 63, 65, 67....

Here a = 63,

=> d = 65-63 = 2

using formula

an =  a + (n-1) 2

=> 63 + (n-1) 2

=> 63 + 2n -2

=> 61 + 2n

2nd AP : - 3,10,17....

Here a = 3, d = 10-3 = 7

using formula

an = a + (n-1) d

=> 3 + (n-1) 7

=> 3 + 7n - 7

=>7n - 4

condition :- nth term of 1st AP = nth term of 2 AP

=> 61 + 2n = 7n -4

=> 61 +4 = 7n - 2n

=> 65 = 5n

65 = n

13 = n

**********************************************************

14) Determine the AP whose third term is "16" and the 7th term exceeds the 5th term by "12"

sol) Third term is "16" :- i.e, a3 = 16,, n = 3

We have :-

an = a + (n-1) d

a3 = a + (3-1) d

a3 = a + 2d

16 = a + 2d ------(1)

7th term :-

a7 = a + ( 7 -1) d

a7 = a + 6d ----(2)

5th term :-

a5 = a + (5-1) d

a5 = a + 4d -----(3)

7th term exceeds the 5th term by "12"

a7 = a5 + 12

a7 - a5 = 12

substitute (2) and (3)

(a + 6d) - ( a + 4d) = 12

a + 6d - a - 4d = 12

2s = 12

d = 12 /2

d = 6

put "d" value in (1) we get

a = 16 - 2d

=> 16 - 2(6)

=> 16 - 12

=> 4

Hence, we got, a = 4, d = 6

a2 = a + d

=> 4 + 6

=> 10

a3 = a2 + d

=> 10 + 6

=> 16

So, the AP is = 4, 10, 12

******************************************************

15) Find the 20th term from the end of the AP : 3, 8,13.....253.

Sol) Here a = 3, d = 8 - 3 = 5

First find which term is 253

Let an= 253

an = a + (n-1) d

253 = 3 + (n-1) 5

253 = 3 + 5n - 5

253 = 5n - 2

253 + 2 = 5n

255 = 5n

255 = n

n = 51

We found "253" is the last 51th term of AP

Now,

We need to find 20th term from the last term.

i.e, (51 - 19)th term

=> 32th term.

We have to find a32

using formula :-

an = a + (n-1) d

a32 = 3 + (32-1)*5

=> 3 + 31 *5

=> 3 + 155

=> 158

So, 20th term from the last term of the given AP is "158"

*********************************************************

16) The sum of 4th and 8th terms of an AP. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.

sol) 4th term :-

an = a + (n-1) d

a4 = a + (4-1)d

a4 = a + 3d ------(1)

8th term :-

an = a + (n-1) d

a8 = a +(8-1)d

a8 = a +7d -------(2)

6th term :-

an = a + (n-1) d

a6 = a +(6-1)d

a6 = a +5d -----(3)

10th term :-

an = a + (n-1) d

a10 = a + (10-1) d

a10 = a +9d -----(4)

Condition (1) :- sum of 4th and 8th term of AP is 24

a4 + a8 = 24

(a +3d ) + (a + 7d) = 24

2a + 10d = 24 -------(5)

condition (2) :- Sum of 6th term and 10th term of AP is 44

a6 + a10 = 44

( a + 5d) + (a + 9d) = 44

2a + 14d = 44

2a = 44 - 14d

a = 44 -14d
2 2

a = 22 - 7d -----(6)

putting (6) in (5) we get

2a + 10d = 24

2(22 - 7d) + 10d = 24

44 - 14d + 10d = 24

44 - 24 -4d = 0

20 = 4d

20 = d
4

5 = d

put "d" value in (6) we get

a = 22 - 7(5)

a 22 - 35

a = -13

We got a = -13,,, d =5

2nd term = 1st term + d

=> -13 + 5

=> - 8

3rd term = 2nd term + d

=> -8 + 5

=> -3

Hence, first three terms are -13, -8, -3

*******************************************************

17) Subha Rao started work in 1995 at an annual salary of 5000 Rs and received an increment of 200Rs each year. In which year did his income reach 7000 Rs?

sol) Salary in 1995(First year) = 5000

Increment of Rs.200 :-

Salary in 1996 ( second year) = 5000 + 200 = 5200

Salary in 1997 ( Third year) = 5200 + 200 = 5400

So, the series is : 5000, 5200, 5400.....

Since difference is same , it is an AP

Common difference (d) = 200

We want to find in which year his income becomes 7000

So, an = 7000

We need to find "n"

Here, an = 7000, a = 5000, d = 200

using formula

an = a + (n-1) d

7000 = 5000 +(n-1) *d

7000 - 5000 = (n-1) * 200

2000 = n - 1

200

10 = n -1

11 = n

.^. , in the 11th year, Subha Rao salary has become 7000

**********************************************************

Example 11) if the sum of the 1st 14terms of an AP is 1050 and its first term is 10, find the 20th term.

Sol) Here

Sn = 1050

n = 14, a = 10

We have

Sn = n [2a +(n-1)d]
   2

1050 = 14 [2a +13d]
   2

1050 = 7 [2a + 13d]

1050 = 14a + 91d

1050 = 14(10) + 91d

1050 - 140 = 91d

910 = 91d

.^. d = 10

.^. a20 = 10 +(20-1)*10

= 200

*****************************************

Example 12). How many terms of the AP : 24, 21,18...... must be taken so that their sum is 78?

sol) Here

a= 24

d = 21 - 24 = -3

Sn = 78.

We need to find "n"

We have

Sn = n [2a +(n-1)d]
   2

78 = n [48 +(n-1)(-3)]
   2

3n^2 - 51n + 156 = 0

n^2 - 17n +52 = 0

(n-4)(n-13) = 0

n = 4 or 13

Both values of "n" are admissible. So, the no.of terms is either "4" or "13"

Remarks :

1) In this case, the sum of the first 4terms = the sum of the first 13 terms = 78

2) Two answers are possible because the sum of the terms from 5th to 13th will be zero. This is because "a" is positive and "d" is negative, so that some terms are positive and some are negative, and will cancel out each other.

*****************************************

Example 13). Find the sum of :

(!) the first 1000 positive integers

sol) Let S = 1 + 2+ 3+....+1000

Using the formula :-

Sn = n (a +L)
   2

for the sum of the first "n" terms of an AP, we have

S1000 = 1000 ( 1+1000)
2

=> 500 * 1001

=> 500500

So, the sum of first 1000 positive integers = 500500

(!!) Let Sn = 1 +2+3+...+n

sol) Here a =1 and the last term"L" is "n"

.^., S = n (1 +n)
                    2

or

S = n(n+1)
             2

So, the sum of first "n" positive integers is given by

Sn = n(n+1)
           2
*****************************************

Example 14) Find the sum of first 24 terms of the list of numbers whose nth term is given by

an = 3 + 2n

sol) a1 = 3 +2(1) => 5

a2 =3 +2(2) => 7

a3 = 3 +2(3) => 9

List of numbers becomes 5,7,9,11....

Here, d1 = 7-5 = 2
d2 = 9-7 = 2
d3 = 11-9 =2 and so on....

So, it forms an AP with common difference d=2

.^., S24 = 24 [ 2*5 +(24-1) *2]
2

=> 12 (10+46)

=> 672.

So, sum of first 24 terms of the list of numbers is 672

******************************************

Example 15) A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increase uniformly by a fixed number every year.

find :

(!) The production in the 1st year

sol) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st, 2nd, 3rd.....years will form an AP.

Let us denote the number of TV sets manufactured in the nth year by an.

Then,

a3 = 600 and a7 = 700

a + 2d = 600

a + 6d = 700

Solving these equations, we get

d=25 and a = 550

.^., the production of TV sets in the first year=550.

(!!) Now

a10 = a +9d

=> 550 + 9*25

=> 775

So, production of TV sets in the 10th year is 775.

(!!!) Also,

S7 = 7 [2*550 + (7-1)*25]
         2

=> 7/2 [ 1100 +150]

=> 4375

Thus, the total production of Tv sets in first 7 years = 4375.

Exercise- 6.3

1) Find the sum of the following APs:

(1) 2,7,12,.....,to 10 terms.

sol) Formula :-

Sum of AP = n { 2a + (n-1)d}

Here n = 10, a = 2,

d = 7-2 = 5

Using formula :-

n { 2a + (n-1)d}

10 {2 * 2 + (10-1) *5

5{4 + 9 *5}

5{4 + 45}

5{49} = 245

******************************************************

(2) -37, -33, -29,... to 12 terms

sol) Here a = -37, n = 12,

d = -33 - (-37)

=> -33 + 37

=> 4

using formula :-

Sum = n { 2a + (n-1)d}

=> 12 { 2 *(-37) + (12-1) *4}

=> 6 { -74 + 11 *4}

=> 6 {-74 + 44}

=> 6{-30)

=> -180

********************************************************

(3) 0.6, 1.7, 2.8,....., to 100 terms

sol) Here a = 0.6, n = 100

d = 1.7 - 0.6 = 1.1

using formula :-

Sum = n { 2a + (n-1)d}

~~100~~ {2 * 0.6 + (100-1) d}

50{ 1.2 + 99 * 1.1}

50 { 1.2 + 108.9}

50 { 110.1}

50 * 110.1

= 5505

*******************************************************

(4) 1 , 1 , 1 ,...., to 11 terms
15 12 10

sol) a = 1 , n = 11
5

d = 1 - 1
12 15

d = 15 -12
12*15

d = 3
180

d = 1
60

using formula :-

Sum = n { 2a + (n-1)d}

=> 11 { 2* 1 + (11-1) * 1 }

2 15 60

=> 11 { 2 + (10) * 1 }

2 15 60

=> 11 { 2 + 10 }

2 15 60

=> 11 { 2 + 1 }

2 15 6

=> 11 { 4 + 5}

2 30

=> 11 ( 9 )

2 30

=> 11 ( 3 )

2 10

=> 33

***********************************************************

2) , Find the sums given below :

(1) 7 + 10 1 + 14 +..... + 84
2

sol) Here a = 7

d = 10 1 - 7
2

d = 21 - 7
2

d = 21 - 14
2

d = 7
2

Also, last term =L = 84

First, we find "n"

an = a + (n-1) d

Here, an = L = 84

84 = 7 + (n-1) 7
2

84 - 7 = (n-1) 7
2

77 * 2 = n-1
7

11 *2 = n-1

22 + 1 = n

n = 23

.^., n =23

We have formula :-

S = n ( a + L)
2

S = 23 ( 7 + 84)
2

S = 2093
2

S = 1046.5

*********************************************************************************

(2) 34 + 32 + 30 +......+10

sol) Here a = 34,

d = 32 - 34 = -2

L = 10

First, we find "n"

an = a + (n-1)d

Here an = L = 10

10 = 34 + (n-1) (-2)

10 -34 = (n-1) (-2)

-24 = (n-1) (-2)

~~-24~~ = n-1
-2

n-1 = 12

n = 12 +1

n = 13

.^., n = 13

So, we use the formula

S = n (a +l)
2

S13 = 13 ( 34 + 10)
2

S13 = 13 ( 44)
2

S13 = 286

********************************************************

(3) -5 + (-8) + (-11) + ......+(-230)

sol) a = -5

d = -8 - (-5)

=>-8 +5

=> -3

Also, last term = L = - 230

First , we find "n"

an = a + (n-1)d

-230 = (-5) + (n-1) (-3)

-230 - (-5) = (n-1) (-3)

-230 +5 = (n-1) (-3)

-225 = (n-1) (-3)

-225 = n-1

-3

n-1 = -225

-3

n-1 = 75

n = 76

.^. , n = 76

So, we use the formula

S = n (a+L)
2

= 76 ( -5 + (-230)

=76 (-235)

= - 8930

*********************************************************

(3) In an AP:

(1) given a= 5, d = 3, an = 50, find "n" and "Sn"

sol) We have

an = a + (n-1)d

50 = 5 + (n-1) *3

50 = 5 + 3n - 3

50 = 2 + 3n

48 = 3n

48 = n

n = 16

Now we need to find Sn

Sn = n { 2a + (n-1) d}

We have n = 16, a = 5, d = 3

= 16 ( 2*5 + ( 16-1) *3 )

=8 (10 + 15*3)

=8 (10 + 45)

= 8 * 55

= 440

************************************************************

(2) given a =7, a13 = 35, find "d" and S13

sol) We know that

an = a + (n-1)d

Finding "d"

Using formula :-

an = a + (n-1)d

a13 = 7 + (13-1)d

35 = 7 + (12) *d

35 - 7 = 12d

28 = 12d

28 = d

7 = d

Now we need to find S13

We can use formula

Sn = n (a +L)

= 13 ( 7 + 35)

= 13 * 42 = 13 *21 = 273

*********************************************

(3) given a12 = 37, d= 3, find "a" and s12.

sol) Finding "a"

Here an = a12 = 37, n =12, d = 3

Using formula :-

an = a + (n-1)d

37 = a + (12-1) *3

37 = a + 11*3

37 = a + 33

37 - 33 = a

a = 4

Finding s12 using formula :-

Sn = n (a +L)

Here n =12, a = 4, L = a12 = 37

s12 = 12 (4 + 37)

=> 6 (41)

=> 246

*****************************************************

(4) given a3 = 15, s10 = 125, find "d" and "a10"

sol) When a3 = 15

a3 = 15 , n = 3

an = a + (n-1)d

a3 = a + (3-1) d

15 = a + 2d ------(1)

Finding s10 :-

s10 = 125, n = 10,

Sn = n (2a + (n-1) d)

s10 = 10 (2a + (10-1)d)

125 = 5 (2a + 9d)

~~125~~ = 2a + 9d

25 = 2a+ 9d --------(2)

Multiplying (eq.1) by "2" , we get

2*(a + 2d = 15)

2a + 4d = 30 ----(3)

Using elimination method:-

2a + 9d = 25

2a + 4d = 30

- - -

0 + 5d = -5

5d = -5

d = -1 -----(4)

Put "d" value in (eq.1)

a + 2(-1) = 15

a -2 = 15

a = 15 +2

a = 17

Now, we got

a = 17, d = -1 , a10=? n = 10

an = a + (n-1)d

a10 = 17 + (10-1) * -1

a10 = 17 + (9) * -1

a10 = 17 - 9

a10 = 8

.^. d = -1 and a10 = 8

**********************************************************

(5) given a = 2, d= 8, Sn = 90, find "n" and an = ?

sol) using formula :-

Sn = n (2a + (n-1) d)

90 = n (2*2 + (n-1)8)

90 *2 = n (4 + 8n - 8)

180 = n (8n -4)

180 = 8n^2 - 4n

8n^2 - 4n - 180 = 0

4(2n^2 - n - 45) = 0

2n^2 -n - 45 = 0

2n^2 - 10n + 9n - 45 = 0

2n(n-5) + 9 (n-5) = 0

(2n+9) ( n-5) = 0

2n +9 = 0 or n - 5 = 0

n = -9 /2 or n =5

But "n" cannot be in fraction,

So, n = 5

Now finding an i.e, a5

using formula :-

here a = 5, d = 8, a5=?, n = 5

an = a + (n-1)d

a5 = 2 + (5-1)8

a5 = 2 + (4)8

a5 = 2 + 32

a5 = 34

.^. we got n = 5 and a5 = 34

********************************************************

(6) Given an = 4, d = 2, Sn = -14, find "n" and "a"

sol) We have :-

Sn = n (a +L)

-14 = n (a + 4)

-14 *2 = n (a +4)

-28 = n (a +4)

-28 = n -------(1)

a +4

finding "a"

an = a + (n-1)d

4 = a + (n-1)2

4 = a + 2n-2

4 + 2 = a + 2n

6 = a + 2n ------(2)

Putting "n" value in above equation.2

6 = a + 2 ( -28 )

a+4

6 = a - 56

a + 4

6 = a(a+4) - 56

a+4

6(a +4) = a^2 + 4a - 56

6a + 24 = a^2 + 4a - 56

a^2 +4a -6a -56 -24 = 0

a^2 -2a - 80 = 0

a^2 - 10a +8a - 80 = 0

a(a - 10) + 8(a - 10) = 0

(a-10) (a +8) = 0

a = 10, or a = -8

When a = 10

n = -28

a + 4

n = -28

10 + 4

n = -28

n = -2

When a = -8

n = -28

a + 4

n = -28

-8 + 4

n = -28

- 4

n = 7

Since "n" is number of terms, it cannot be negative

.^. a = 10 is not possible

Hence, n= 7, and a = -8

*******************************************************

(7) given L =28, S = 144, and there are total 9 terms. Find "a".

sol) Given L = 28, S = Sn = 144, n = 9

Sn = n (a +L)

144 = 9 (a + 28)

144 * 2 = a + 28

288 = a +28

32 - 28 = a

a = 4

********************************************************

(4) The first and the last terms of an AP are 17 and 350 respectively. If the common difference is "9", how many terms are there and what is their sum?

sol) Given :- a = 17, L = 350, d = 9, n=? , Sn=?

Sn = n (a +L)

Sum = n (17 + 350)

Sum = n * 367 ------(1)

ALSO,

using formula :-

Sn = n (2a + (n-1) d)

2

Sum = n { 2 * 17 + (n-1) * 9}
2

Sum = n { 34 + 9n - 9 }
2

Sum = n { 25 + 9n} ---------(2)
2

Equating (1) and (2)

367 * n = n {25 + 9n}
2 2

367 = 25 + 9n

367 - 25 = 9n

342 = 9n

342 = n
9

38 = n

Put "n" value in (1)

Sum = 367 * 38
2

=> 367 * 19

=> 6973

********************************************************

(5) Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

sol) Given 2nd term = 14

a2 = a + (2-1) d

14 = a + d

14 - d = a --------(1)

Given 3rd term = 48

a3 = a + (3-1)d

18 = a + 2d

18-2d = a ------(2)

Equating (1) and (2)

14 - d = 18 - 2d

2d - d = 18 - 14

d = 4

Putting "d = 4" in (1)

a = 14 -d

a = 14 -4

a = 10

Now, we need to find sum of first 51 terms

Using formula

Sn = n (2a + (n-1) d)

2

s51 = 51 { 2 * 10 + (15-1) * 4 }
2

s51 = 51 { 20 + 50 *4}
2

s51 = 51 {20 + 200}
2

s51 = 51 * ~~220~~
2

s51 = 51 * 110

s51 = 5610 is the sum of first 51 terms

************************************************************

6) If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first "n" terms.

sol) Sum of first 7 terms (s7)= 49

Here n = 7

Sn = n (2a + (n-1) d)

2

49 = 7 (2a + ( 7-1)d)
2

49 = 7 (2a + 6d)
2

49 * 2 = 2a + 6d
7

7 * 2 = 2a + 6d

14 - 6d = 2a

14 - 6d = a
2

14 - 6d = a
2 2

7 - 3d =a -----(1)

Sum of first 17 terms (s17) = 289

n = 17

Sn = n (2a + (n-1) d)

289 = 17 ( 2a + (17-1) d)

289 * 2 = 2a + 16d

~~578~~ = 2a + 16d

34 = 2a + 16d

34 - 16d = 2a

34 - 16d = a

2 2

17 - 8d = a -------(2)

equating (1) and (2)

7 - 3d = 17 - 8d

8d - 3d = 17 -7

5d = 10

d = 10

d = 2

put "d = 2" in (eq.1)

a = 7 - 3d

a = 7 - 3*2

a = 7 - 6

a = 1

.^. a = 1 & d = 2

Finding the sum of first "n" terms :-

Using Formula :-

Sn = n (2a + (n-1) d)

=> n (2*1 + (n-1)2

=> n (2 + 2n -2)

=> n ( 0 + 2n)

=> n * 2n

=> n *n

=n^2

***************************************************************

(7) Show that a1, a2, .....an,....form an AP where an is defined as below

(1) an = 3 + 4n. Also find the sum of the first 15 terms in each case.

sol) Given :- an = 3 + 4n.

When n =1

a1 = 3 + 4*1

a1 = 3 + 4

a1 = 7

When n = 2

a2 = 3 + 4 *2

a2 = 3 + 8

a2 = 11

When n = 3

a3 = 3 + 4*3

a3 = 3 + 12

a3 = 15

Here the series is 7, 11, 15....

Common difference :-

(d1) = 11-7 = 4

d2 = 15 - 11 = 4

Since d1 = d2 ..it is an AP.

Finding sum of first 15 terms i.e, s15 :-

We have a =7, d = 4, n = 15, s15=?

Sn = n (2a + (n-1) d)

2

=>15 ( 14 + 14*4)
2

=> 15 ( 14 + 46 )
2

=> 15 ( 70)
2

=> 15 * 35

= 525

*****************************************************

(2) an = 9 - 5n Also find the sum of the first 15 terms in each case.

sol) Given :- an = 9 - 5n

When n = 1

a1 = 9 -5*1

a1 = 9-5

a1 = 4

When n=2

a2 = 9 - 5*2

a2 = 9- 10

a2 = -1

When n = 3

a3 = 9 - 5*3

a3 = 9 - 15

a3 = -6

Here the series is: 4, -1, -6,....

d1 = -1-4 = -5

d1 = -6 + 1 = -5

Since d1 = d2 . It is an AP

Finding sum of first 15 terms i.e, s15 :-

Here a = 7, d = 4, n = 15, s15=?

Sn = n (2a + (n-1) d)

2

s15 = 15 ( 2 *4 + (15-1) * -5)
2

=> 15 ( 8 + 14 * -5 )
2

=> 15 ( 8 - 70)
2

=> 15* - 62
2

=> 15 * (-31)

= - 465
*********************************************************

(8) If the sum of the first "n" terms of an AP is 4n - n^2, what is the first term ( remember the first terms is s1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

sol) Given Sn = 4n - n^2

Taking n = 1

s1 = 4 * 1 - 1^2

=> 4-1 = 3

SO, sum of first term of AP is 3

But sum of first term will be the first term

So, First term (a) = 3

Taking n = 2 in Sn

s2 = 4 *2 - 2^2

=> 8 - 4

=> 4

.^. Sum of first 2 terms is 4

Finding 2nd term :-

Sum of first 2 terms is 4

First term + Second term = 4

3 + a2 = 4

a2 = 4 - 3

a2 = 1

Hence in AP,

First term (a) = 3

common difference (d) = 2nd term - 1st term

=> 1 -3

=> -2

We need to find 3rd, 10th and the nth term

We know that

an = a (n-1) d

Third term :-

a3 = a + (3 -1) d

=> a + 2d

=> 3 + 2* (-2)

=> 3 - 4

=> -1

10th term :-

a10 = a + (10-1) d

=> a + 9d

=> 3 + 9 * -2

=> 3 - 18

=> -15

nth term :-

an = a + (n-1) d

=> 3 + (n-1) * (-2)

=> 3 + n (-2) -1 * (-2)

=> 3 - 2n +2

=> 5 - 2n

****************************************************

(9) Find the sum of the first 40 positive integers divisible by "6"

sol) Positive integers divisible by 6:-

6/6 =1 , 12/6 = 2, 18,6 = 3

40 positive integers divisible by 6 :- 6, 12, 18,.........

Since difference is same, it is an AP.

Finding sum of 40 integers :-

Using formula :-

Sn = n (2a + (n-1) d)

2

Here, n = 40, a = 6, d =12-6 = 6

Sn = 40 (2*6 + (40-1) * 6)
2

Sn = 20 (12 + 39 *6)

Sn = 20( 12 + 234)

Sn = 20 * 246

Sn = 4920

.^., the sum of first 40 multiples of 6 is 4920

********************************************************

(10) A sum of 700Rs is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is 20rs less than its preceding prize, find the value of each of the prizes.

sol) Let the 1st prize be "a"

2nd prize = a - 20

3rd prize = (a-20) - 20 = a - 40

4th prize = (a-40) - 20 = a - 60

And so on till 7th prize.

So, the series will be :- a, a -20, a-40, a-60,....

Since the difference is same, it is an AP

Here, sum of Rs 700 is given,

So, Sn = 700

Here a = a , d = a-20-a = -20, n = 7

using formula :-

Sn = n (2a + (n-1) d)

2

700 = 7 (2a + (7-1) * (-20))
2

700 = 7 (2a + 6 * (-20)
2

700 = 7 (2a - 120)
2

~~700~~ *2 = 2a - 120
7

100 * 2 = 2a - 120

200 = 2a - 120

200 +120 = 2a

320 = 2a

~~320~~ = a
2

a= 160

So,

1st prize (a) = 160

2nd prize(a-20) = 160 -20 = 140

3rd prize (a - 40) = 160-40= 120

4th prize (a - 60) = 160 - 60 = 100

5th prize ( a-80) = 160 - 80 = 80

6th prize ( a - 100) = 160 - 100 = 60

7th prize ( a -120) = 160 - 120 = 40

*********************************************************

11) In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees. that each section of each class will plant, will be same as the class, in which they are studying. e.g, a section of Class 1 will plant 1 tree, a section of Class-2 will plant 2 trees and so on till Class 9. There are three sections of each class. How many trees will be planted by the students?

sol) condition:- There are 3-section for every class and only one 1 tree is allowed for class1, only 2 trees allowed for class2, similarly till class9.

.^. Trees planted by class 1st = 1(tree) * 3(sections)

=> 1 *3 = 3

Trees planted by class 2nd = 2 * 3 = 6

Trees planted by class 3rd = 3 * 3 = 9

So on....

Trees planted by class 12th = 12 * 3 = 36

Hence the series is :- 3, 6,9,.....36

Since difference is same, it is an AP

finding total number of trees planted :-

Since we have first a = 3 and last value = 36. We can use this formula

using Formula :-

Sn = n (a +L)
2

Here n = 12, a = 3, L = 36

Sn = 12 (3 + 36)
2

=> 6 * 39

=> 234

Hence the total number of trees planted by school is 234

***************************************************************

12) A spiral is made up of successive semicircles, with centers alternately at A and B, starting wit center at A. of radii 0.5cm, 1.0cm, 1.5cm, 2.0cm,..... as shown if figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?

( Take "pi" = 22/ 7)

{ Hint:- Length of successive semicircles is L1, L2, L3, .....with centres at A, B, A, B,....., respectively, }

sol) Here we have to find total length i.e, total circumference of all semicircles

we know that

Circumference of circle = 2piR

So, circumference of semicircle = 1/2 * 2piR = PiR

Circumference of 1st semi-circle = Pi * 0.5

Circumference of 2nd semicircle = Pi * 1.0

Circumference of 3rd semicircle = Pi * 1.5

So on... til 13th semicircle

Circumference of 13th semicircle = Pi * 6.5

So, the series is

Pi * 0.5, Pi * 1.0, Pi * 1.5.......Pi * 6.5

Taking "Pi" as common

= Pi ( 0.5 , 1.0, 1.5,....6.5)

= Pi (S)

As "S" have same difference , it is an AP

"S" can be found using formula

Sn = n (a +L)

2

Here, a = 0.5, L = 6.5, n = 13

S = 13 ( 0.5 + 6.5)
2

S = 13 * 7
2

S = 91
2

S = 45.5

length of spiral = Sum of series

=> Pi * S

=> Pi * 45.5

=> 22 * 45.5
7

=> 22 * 6.5

=> 143

.^. length of spiral is 143 cm

****************************************************************

13). 200 logs are stacked in the following manner. 20 logs in the bottom row, 19 in the next row, 18 in the row next to it so on. In how many rows are the 200 logs placed and how many logs are in the top row?

sol) Let a = 20

d= 19-20 = -1

Since difference is same in each row , it is an AP

using formula :-

Sn = n (2a + (n-1) d)

200 = n ( 2(20) + (n-1) (-1))
2

200 = n ( 40 -n +1)
2

200 = n (41 -n)
2

400 = 41n - n^2

n^2 - 41n + 400 = 0

n^2 -25n -16n + 400 = 0

n(n-25) -16(n - 25) = 0

n-25 = 0 or n -16 = 0

n = 25 or n = 16

But "n" cannot be "25" , Since a = 20,

so n= 16

.^. there are total 16 logs are in the top row.

******************************************************************

(14) In a bucket and ball race, a bucket is placed at the starting point, which is 5m from the first ball, and the other balls are placed 3 m apart in a straight line. There are ten balls in the line.

A competitors starts from the bucket , picks up the nearest ball, rund back with it drops it in the bucket, run back to pick up the next ball, rund to the bucket to drop it in, and she continues in the same way until all the balls are in the bucket. What is the total distance the competitor has to run?

{ Hint : To pick up the first ball and the second ball, the total distance (in meters) run by a competitor is 2*5 + 2* (5+3)}

sol) Distance run to pick 1st potato = 5m each side

=> 5m * 2

=> 10m both side

(*) Distance run to pick second potato = 5 + 3m each side

=> 8m each side

= 16m both sides

(*) Distance run to pick third potato = 8 + 3m each side

=> 11m each side

=> 22m both side

Hence the series is 10, 16, 22

Since the difference is same, it is an AP

Now, we need to calculate total distance she runs to pick 10 potatoes

i.e, s10

using formula :-

Sn = n (2a + (n-1) d)

         2

Here, n =10, ( 10 potatoes)

a = 10, d = 16-10= 6

Sn = 10 (2 *10 + (10-1) * 6)
2

=> 5 (20 + 9*6)

=> 5 (20 +54)

=> 5 * 74 = 370

Hence, total distance covered = 370m

*********************************************************

Example 16) Write the GP. if the first term a= 3, and the common ratio r =2.

sol) Since "a" is the first term it can easily be written

We know that in GP, every succeeding term is obtained by multiplying the preceding term with common ratio"r". So to get the second term we have to multiply the first term a=3 by the common ratio r = 2

.^. Second term = a*r = 3*2 =6

Similarly the third term = 2nd term * common ratio

=> 6*2 = 12

If we proceed in this way we get the following G.P.

3,6,12,24.........

*****************************************

Example 17) Write GP. if a = 256, r =(-1/2)

sol) general form of GP = a, ar, ar^2,ar^3.......

=> 256, 256(-1/2), 256(-1/2)^2, 256(-1/2)^3

=> 256, -128, 64, -32.....

******************************************

Example 18) find the common ratio of the GP 25, -5, 1, (-1/5)

sol) We know that if the first, second, third....terms of a GP are a1, a2,a3....respectively

the common ratio:-

r = a2/a1 = a3/a2=......

Here a1 = 25, a2 = -5, a3=1

So common ratio

r = (-5/25)

=> - (1/5)

*****************************************

Example 18) Which of the following list of numbers form GP?

(!) 3,6,12,-------

sol) We know that a list of numbers a1,a2,a3......an....is called a GP if each term is non zero and

a2 = a3 = .....a^n = r
a1 a2 a^(n-1)

Here all the terms are non-zero. Further

a2 = 6 = 2   and
a1    3

a3 = 12 = 2
a2    6

a2 = a3 = 2
a1 a2

So, the given list of number form a G.P. which contain ratio 2.

(!!) (1/64), (1/32), (1/8),....

sol) All the terms are non-zero

a2 = (1/32) = 2
a1 (1/64)

a3 = (1/8) = 4
a2 (1/32)

Here a2 =/= a3
   a1        a2

So, the given list of numbers does not form GP.

******************************************

Exercise - 6.4

1) In which of the following situations, does the list of numbers involved in form a GP?

(1) Salary of Sharmila, when her salary is 5,00,000 for the first year and expected to receive yearly increase of 10%

sol) It is not a GP

Given :- Sharmila's salary for the first year = 5,00,000

Yearly increase = 10 * 5,00,000 = 50,000
~~100~~

Thus, the salary for the second year

=> 5,00,000 + 50,000 = 5,50,000

Similarly

salary for 3rd year = 5,50,000 + 50,000 => 6,00,000

ratio between 2nd and 1st year salary

=> ~~5,50,000~~ = 1.1 -----------(1)
5,00,000

ratio between 3rd and 2nd year salary

=> ~~6,00,000~~ = 1.09 -------(2)
5,50,000

Since the ratio is not same it is not a GP.

*******************************************************

(2) Number of bricks needed to make each step, if the stair case has total 30 steps. Bottom step needs 100 bricks and each successive step needs 2 bricks less than the previous step.

sol) If the bottom step needs 100 bricks,

the subsequent step needs "2" bricks less than the bottom step.

the step next to bottom step needs ="98" bricks.

The 2nd last step needs = 96 bricks

Thus, the series is :- 100, 98.....42

1) Ration between second and first step

= 98 = 0.98
100

2) ratio between 3rd and 2nd step

= 96 = 0.979
98

Since the ratios are not equal, it is not a GP.

**********************************************************

(3) Perimeter of the each triangle, when the mid-points of sides of an equilateral traingle whose side is 24 cm are joined to form another triangle, whose mid points in turn are joined to form still another triangle and the process continues indefinitely.

sol) Perimeter of an equilateral triangle,
with side 24 cm = 3 * 24 = 72 cm -----(1)

Perimeter of 2nd inner traingle.
with side 12cm = 3 *12 = 36 cm -----(2)

Perimeter of 3rd inner triangle.
with side 6 cm = 3 * 6 = 18 cm ......(3)

Ratio of the perimeter of the 2nd and 1st triangle
= 36 = 1 ----(4)
72 2

Ratio of the perimeter of 3rd and 2nd triangle.

=> 18 = 1 -----(5)
36 2

Ratios of the triangles are same. Hence it is a GP

**************************************************************

(2) Write three terms of the GP when the first term "a" and the common ration "r" are given?

(1) a = 4; r =3

sol) general form of GP

=> a, ar, ar^2, ar^3,...

=> 4, 4(3), 4(3)^2, 4(3)^3

=> 4, 12, 36, 108...

**************************************

(2) a = _/5; r = 1/5

sol) general form of GP

=> a, ar, ar^2, ar^3....

=> _/5, _/5(1/5), _/5(1/2)^2, _/5(1/2)^3.....

=> _/5, _/5/5, _/5 / 25

***************************************

(3) a = 81; r = -1/3

sol) a, ar, ar^2, ar^3...

=> 81, 81(-1/3), 81(-1/3)^2....

=> 81, -81/3, -81/9

=> 81, -27, -9

***************************************

(4) a = 1/ 64 ; r = 2

sol) a, ar, ar^2, ar^3...

=> 1/64, 1/64(2), 1/64(2)^2, 1/64(2)^3

=> 1/64, 1/32, 1/16, 1/8...

**************************************

(3) Which of the following are GP? If they are GP. Write three more terms?

(1) 4, 8, 16......

sol) We know that a list of numbers a1, a2, a3..... is called a GP if each term is non-zero and have same ration

i.e, a2 = a3 = r
a1 a2

Here, a2 = 8 = 2
a1 4

a3 = 16 = 2
a2 8

i.e, a2 = a3 = 2
a1 a2

Thus the common ration is "2"

So, the given list of the numbers forms a GP with the first term as a = 4 and the common ration r =2. Thus the next three terms are :

a =4, ar = (4*2) =8,

ar^2 = 4(2)^2 = 16

ar^3 = 4(2)^3 = 32,

ar^4 = 4(2)^4 = 64

ar^5 = 4(2)^5 = 128

***********************************

(2) 1/3, -1/6, 1/12......

sol) Here a2 = -1/6 = -1
a1 = 1/3 2

a3 = 1/12 = -1
a2 = -1/6 2

a2 = a3 = -1   is the common ratio.
a1 a2 2

a = 1/3,

ar = 1/3(-1/2) = -1/6

ar^2 = 1/3(-1/2)^2 = 1/12

ar^3 = 1/3(-1/2)^3 = -1/24,

ar^4 = 1/3(-1/2)^4 = 1 /48

ar^5 = 1/3(-1/2)^5 = -1/ 96

*********************************

(3) 5, 55, 555....

sol) Here a2 = 55 = 11
a1 5

a3 = ~~555~~ = 10.09
a2 55

Since a2 =/= a3
a1 a2

So, the given list of the numbers does not form a GP

********************************************

(4) -2, -6, -18,....

sol) a2 = -6 = 3
a1 -2

a3 = -18 = 3
a2 -6

Since a2 = a3 = 3,
a1 a2

We have a = -2, r = 3,

ar = -2(3) = -6

ar^2 = -2(3)^2 = -18

ar^3 = -2(3)^3 = -54

ar^4 = -2 (3)^4 = -162

ar^5 = -2(3)^5 = -486

*************************************

(5) 1/2, 1/4, 1/6...

sol) a2 = 1/4 = 1
a1 1/2 2

a3 = 1/6 = 2
a2 1/4    3

Since a2 =/= a3
a1 a2

So, the given list of the numbers does not form a GP.

**********************************************

(6) 3, -3^2, 3^2....

sol) a2 = -3^2 = -3
a1 3

a3 = 3^3 = -3
a2 = -3^2

Since a2 = a3 = -3
a1 a2

We have a = 3,, r = -3

ar = 3(-3) = -9

ar^2 = 3(-3)^2 = 27

ar^3 = 3(-3)^3 = -81

ar^4 = 3 (-3)^4 = 243

ar^5 = 3(-3)^5 = -729

********************

(7) x , 1 , 1/x....

sol)  a2 = 1
a1    x

a3 = (1/x) = 1
a2 1 x

Since, a2 = a3 = 1
a1 a2 x

We have a = x, r = (1/x)

ar = x*(1/x) = 1

ar^2 = x * (1/x)^2 = 1/x

ar^3 = x * (1/x)^3 = 1 / (x^2)

ar^4 = x * (1/x)^4 = 1 / (x^3)

ar^5 = x* (1/x)^5 = 1 / x^4

**************************

(8) 1 / _/2 , -2, 8 / _/2....

sol) a2 = -2 = -2_/2
a1 (1/_/2)

a3 = 8_/2 = -4 = -4 * _/2 = -4 * _/2 = -2_/2
a2 -2 _/2 _/2 * _/2 2

Since a2 = a3 = -2 _/2
a1 a2

Here a = 1 / _/2, r = -2_/2

ar^3 = 1 / _/2 * (-2_/2)^3 = -16

ar^4 = (1 / _/2) * ( -2/_/2) ^4 = 32_/2

ar^5 = (1/_/2) * ( -2_/2)^5 = -128

*********************************

(9) 0.4, 0.04, 0.004......

sol) a2 = 0.04 = 0.1
a1 0.4

a3 = 0.004 = 0.1
a2 0.04

Since a2 = a3 = 0.1
a1 a2

We have a = 0.4,, r = 0.1

ar^3 = 0.4*(0.1)^3 = 0.0004

ar^4 = 0.4*(0.1)^4 = 0.00004

ar^5 = 0.4*(0.1)^5 = 0.000004

****************************

(4) Find "x" so that x, x +2, x +3 are consecutive terms of a geometric progression.

sol) Given :- the series is in GP

We have , a1 = x, a2= (x+2), a3 = (x +3)

i.e, a2 = a3
a1 a2

x + 2 = x +3
x x+2

(x+2)^2 = x(x+3)

x^2 + 4x + 4 = x^2 + 3x

4x + 4 = 3x

x + 4 = 0

x = -4

***************************************

Example 20) Find the 20th and nth term of the GP.

5, 5, 5....
2 4 8

sol) Here

a = (5/2)

r = (5/4) / (5/2)

=> 1/2

Then

a20 = ar^n-1

=> ar^20-1

=> (5)*(1/2)^ 19
   2

=>     5
2^10

and

a = ar^n-1

=> 5 * (1/2)^n-1
   2

=>   5
2^n

******************************************

Example 21) Which term of the GP: 2, 2√2, 4.....is 128?

sol) Here

a = 2,

r = 2√2
2

=> √2

Let 128 be the nth term of the GP.

Then,

a = ar^n-1

128 = 2(√2)^n-1

(√2)^n-1 =128 = 64

   2

(2)^(n-1) = 2^6
2

(n-1) = 6
   2

.^. n = 13

Hence 128 is the 13th term of the GP.

*****************************************
Example 22). In a GP the 3rd term is 24 and 65th term is 192. Find the 10th term.

sol) Here

a3 = ar^2 = 24 -------(1)

a6 = ar^5 = 195 ------(2)

Dividing (2) by (1) we get

ar^5 = 192
ar^2    24

=> r^3 = 8 = 2^3

=> r = 2

substituting r=2 in (1) we get a =6

.^. a10 = ar^9

=> 6(2)^9

=> 3072

******************************************

Exercise-6.5

(1) For each geometric progression find the common ratio "r" and then find an.

1) 3, 3/2, 3/4, 3/8,....

sol) Here a = 3,

r = (3/2) = 1/2
3

an = ar^(n-1)

=> 3( 1/2) ^(n-1)

***************

(2) 2, -6, 18, -54

sol) Here a = 2,

r = (-6) / 2 = -3

an = ar^(n-1)

=> 2(-3)^(n-1)

*****************

(3) -1, -3, -9, -18,....

sol) here a = -1

r = (-3) / (-1) = 3

an = ar^(n-1)

=> -1(3)^(n-1)

*******************

(4) 5,2, 4/5, 8/25....

sol) Here a = 5

r = 2 /5

an = ar^(n-1)

=>5(2/5) ^(n-1)

********************

2) Find the 10th and nth term of GP : 5, 25, 125,.....

sol) Here a = 5 and r = 25/5 = 5

an = ar^(n-1)

a10 = 5(5)^10-1 => 5(5)^9 => 5^10

an= ar^(n-1) = 5(5)^(n-1) => 5^n

***********************************

(3) Find the indicated term of each geometric Progression

1) a1 = 9; r = 1/3; find a7

sol) an = ar^(n-1)

a7 = 9( 1 )^(7-1)

=> 9( 1 ) ^6

=> 3^2

3^6

=> 1

3^4

****************************

2) a1 = -12; r = 1/3; find a6

sol) an = ar^(n-1)

a6 = -12( 1 )^ (6-1)

=> -12 ( 1 )

3^5

=> -12 * ( 1 )

3 3^4

=> -4 * ( 1 )

3^4

=> -4

3^4

*************************

(4) Which term of the GP.

1) 2,8, 32,..... is 512?

sol) Here a = 2,

r = 8 = 4
2

Let "512" be the nth term of the GP.

an = 512

an = ar^(n-1)

512 = 2(4) ^(n-1)

512 = (4) ^(n-1)
2

256 = 4^(n-1)

4^4 = 4^(n-1)

4 = n-1

n = 5

Hence 512 is the 5th term of the GP.

***************************************

2) _/3, 3, 3 _/3..... is 729?

sol) Let 729 be the nth term of the GP.

an= 729, a = _/3

an = ar^(n-1)

729 = _/3 (_/3) ^(n-1)

729 = (_/3) ^n-1+1

3^6 = (_/3)^n

3^6 = (3) ^n/2

6 = n / 2

12 = n

Hence, 729 is the 12th term of the GP.

*****************************************

3) 1/3, 1/9, 1/27...... is 1 / 2187 ?

sol) Let, 1 be the nth term of the GP.
2187

an = ar^(n-1)

1 = 1 ( 1 )^n-1

2187 3 3

1 = ( 1 )^n

3^7 3

1 = 1^n

3^7 3^n

3^n = 3^7

n = 7

Hence, 1 is the 7th term of the GP.

2187

*************************************

(5) Find the 12th term of a GP. whose 8th term is 192 and the common ratio is "2"

sol) Given :- 8th term is 192

an = ar^(n-1)

a8 = ar^(8-1)

192 = ar^7

Given :- r =2

192 = a(2)^7

192 = a(128)

192 = a

128

3 /2 = a

So,

an = ar^(n-1)

a12 = 3 (2)^(12-1)

=> 3 (2)^11

=> 3 *(2)^11 * 2^-1

=> 3 * (2)^10

=> 3072

*********************************

(6) The 4th term of a geometric progression is 2/3 and the 7th term is 16/ 81. Find the geometric series.

sol) Given :- 4th term of a GP = 2/3

an = ar^(n-1)

a4 = ar(4-1)

2 = ar^3 -------(1)

Given :- 7th term of a GP = 16 / 81

an = ar^(n-1)

an = ar^(7-1)

16 = ar^6 ------(2)

Dividing (2) by (1) we get

ar^6 = (16/81) = 8

ar^3 (2/3) 27

r^6-3 = 8

r^3 = 2^3

3^3

r^3 =( 2 )^3

r = 2

Put "r = 2/3" in (1) ..we get

2 = ar^3 -------(1)

2 = a ( 2/3)^3

a = (3/2)^2

a = 9/4

.^. GP are

a = 9/4

ar = 9/4 * (2/3) = 3/2

ar^2 = 9/4 (2/3)^2 = 1

so on...

************************************************

(7) if the geometric progressions 162, 54, 18.... and 2/ 81, 2/27, 2/9.... have their nth term equal. Find the value of "n"

sol) Case (1) :- GP :- 162, 54, 18.....

here a = 162

r1 = 54 = 1

162 3

Case (2) :- GP :- 2/81, 2/27, 2/9,....

Here b = 2/81,

r2 = (2/27) / (2/81) = 3

Given nth term of Case(1) and case(2) are equal

ar1^(n-1) = br^(n-1)

~~162~~ ( 1/3)^n-1 = 2 ( 3)^n-1

81 (1/3)^n-1 = 1 (3)^n-1

3^4 (1/3)^n-1 = 1 (3)^n-1

3^4

(1)^n-1 = 1 (3)^n-1

3 3^8

( 1 )^n-1 = ( 1 )

3 3^8-n+1

n-1 = 8+1-n

n-1 = 9 -n

2n = 10

n = 5

***************************************

Optional Exercise:-

1) Which term of the AP : 121,117,113,......is the first negative term?

[Hint : Find "n" for an<0]

sol) we got [Hint : Find "n" for an <0]

Lets find "an"

Let the nth term of the AP be its first negative term

i.e, an<0

a +(n-1)d <0

Here a= 121 and d= 117-121 =-4

121 +(n-1)*(-4) < 0

121 - 4n +4 < 0

125 - 4n < 0

125 < 4n

125 < n
4

31.25 < n

Since, the first negative term is greater than 31.25,

It is 32nd term a negative integer.

***************************************

2) The sum of the third and the seventh terms of an AP is 6 and their products is 8. Find the sum of first sixteen terms of the AP.

sol) Condition 1:- sum of 3rd and 7th terms of an AP, i.e,

a3 + a7 = 6 ------(1)

Condition 2 :- product of 3rd and 7th term

a3 * a7    = 8 ------(2)

R.T.P :- Sum of first sixteen terms S16 = ?

Now,

a3 = a +(3-1)d => a +2d ----(3)

a7 = a+(7-1)d => a +6d ----(4)

Substitute(3) & (4) in (1)

We get,

Sum of 3rd and 7th term= 6

a + 2d + a +6d = 6

2a +8d = 6

2(a +4d) = 6

a + 4d = 3

or

a = 3-4d -----(5)

Product of 3rd and 7th term = 8

a3 * a7 = 8

(a + 2d) * (a +6d) = 8

****************
a = 3-4d from (5)
****************

(3-4d+2d) * (3-4d+6d) =8

(3-2d)(3+2d) =8

*********************
(a+b)*(a-b) = a^2 - b^2
*********************

(3)^2 - (2d)^2 = 8

9 - 4d^2 = 8

9-8 = 4d^2

1 = 4d^2

1 = d^2
4

√(1/4) = d

1 = d ----------(6)
2

Put (6) in (5), we get

a = 3 - 4(1/2)

a = 3 - (4/2)

a = 6 -4
        2

a = 2/2 = 1 ---------(7)

Thus, sum of 16th term is given by:-

S16 = (n/2) [ 2a + (n-1)d]

substitute "a" and "d" value

we get,

= 16 [ 2*(1) + (16-1)*1/2]
2

= 8 [ 2 + (15)*1/2]

= 8 [ 2 + (15/2)]

= 8 [ 2 + 7.5]

= 8 * 9.5

= 76.0

Hence, sum of 1st 16 terms S16 = 76

*****************************************


(3) A ladder has rungs 25 cm is apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. if the top and bottom rungs are 2(1/2) m apart, what is the length of the wood required for the rungs?

[Hint : Number of rungs = 250 +1]
25

Sol) Given : Distance between 2 rungs =25cm

Total length = 2(1/2)m

=> 4+1 m
   2

=> 5 m
   2

convert "m" into "cm"

100cm = 1m

=> 5 * 100
   2

=> 500 = 250 cm
2

Total Length = 250 cm

Given:- length of the 1st rung(a) = 45 cm

now,

No.of rungs :-

=     Total length      + 1

   Distance between 2 rungs

=    250    + 1
25

= 10 + 1

No. of rungs = 11

Here, length of the rung increases from 25 cm to 45 cm

So, AP is 25........45

Where a = 25

Last term(L) = 45

No.of terms or rungs = 11

R.T.P :- Total wood required for rungs

Total wood required for rungs = Sum of "n" terms of AP

We use the formula

S = n (a +L)
2

Putting n= 11, a =25, L= 45

S = 11 (45 +25)
   2

=> 11* (70)
2

=> 11 * 35

Thus, 385 cm is the length of the wood required

******************************************

4) The house of a row are numbered consecutively from 1-49. Show that there is a value of "x" such that the sum of the numbers of the houses preceding the house numbered"x" is equal to the sum of the numbers of the houses following it. Find this value of "x".

[Hint : Sx-1 = S49 - Sx ]

sol) Using Hint we can find the value of "x"

We have:-

Sn = n(n+1)
2

Sx-1 = (x-1) (x-1+1)
   2

=> (x-1)(x)
   2

=> x^2 - X -------(1)
2

S49 = 49(49+1)
2

=> 49(50)
2

=>2450    ---------(2)
2

Sx = x(x+1)
   2

=> x^2 + X ---------(3)
   2

Now,

Substitute (1),(2) and (3) in

Sx-1 = S49 - Sx

we get,

x^2 - x = 2450 - x^2 - x
2 2 2

x^2 - x = 2450 - x^2 - x

2x^2 - x +x = 2450

2x^2 = 2450

x^2 = 1225

x= √1225                    5 1225
                                   5   245
x = √5^2 *√7^2         7   49
                                   7 7
x = 5 * 7                         1

x = 35

*******************************************

(5) A small terrace at a football ground comprises of 15 steps each of which is 50m long and built of solid concrete. each step has a rise of (1/4)m and a tread of (1/2)m.

Calculate the total volume of concrete required to build the terrace.

[ Hint : volume of concrete required to build the first step = (1/4) * (1/2)*50m^3]

sol) Given:-

No. of steps(n) = 15

length of steps(L) = 50m

Rise in each step(H) =(1/4)m

Tread(breadth) of each step(b) = (1/2)m

Total volume of concrete required = sum of the volumes of individual steps.

Observe each step is a cuboid.

*********************
Volume of cuboid:-

V = Base area * height

= Length * Breadth * Height

= L * b * h

***************************

using Hint

(*) Volume of 1st step :-

=> L * b * h

=> 50 * (1/4) * (1/2) m^3

=> 50 m^3
8

=> 6.25 m^3

(*) height of 2nd step :-

height = 2 * height of 1st step

=> 2 * (1/2) m

=> 1 m

(*) Volume of 2nd step :-

=> L * b * h

=> 50 * (1/4) * 1

=> 50
4

=> 12.5 m^3

(*) height of 3rd step :-

height = 3 * height of 1st step

=> 3 * (1/2)

=> (3/2) m

(*) Volume of 3rd step :-

=> L * b * h

=> 50 * (1/4) * (3/2)

=> 150
   8

=> 18.75 m^3

We got the series ;

6.25, 12.5, 18.75........

First term(a) = 6.25

d1 = a2 - a1 = 12.5 - 6.25 = 6.25

d2 = a3 - a2 = 18.75 -12.5 = 6.25

.^. the volume is in AP

Now,

total volume of concrete = sum of volumes of individual steps(s15)

We Have

Sn = n [2a + (n-1)d]
2

n = 15, a = 6.25, d = 6.25

S15 = 15 [(2*6.25) + (15-1)*(6.25)]
   2

****************************
2*6.25 = 12.5,   14 * 6.25 = 87.5

***************************

=> 15 [ 12.5 + 87.5]
   2

=> 15 * [ 100]
   2

=> 15 * 50

Thus, 750 m^2 is the total volume of the concrete required

*****************************************

(6) 150 workers were engaged to finish a piece of work in a certain no.of days. Four workers dropped from the work in the 2nd day. Four workers dropped in 3rd day and so on...It took 8 more days to finish the work. Find the number of days in which work was completed.

sol) Given :- It takes 8 more days if 4workers dropped in each day.

Assume:-

150 workers can finish the work in (n-8) days, if all workers work all the days.

Then,

Total work = 150(n-8) ----(1)

150 workers on day-1

146 workers on day-2

142 workers on day-3 ......

.^. Total work = 150 + 146+142...n

This is in AP with

a =150, d = 146-150 =-4

.^. Total work :-

Sn = n [ 2*150 +(n-1)*(-4)]
         2

=> n [ 300 -4n +4]
      2

=> n (152 -2n) ------(2)

equate (1) and (2) we get,

150(n-8) = n(152 -2n)

150(n-8) = n*2(76 -n)

75(n-8) =n (76-n)

75n - 600 = 76n - n^2

n^2 - n - 600 = 0

n^2 - 25n + 24n -600 =0

n(n-25) + 24(n - 25) = 0

(n+24)(n-25) = 0

n= -24 n=25

since days cannot be in -ve

Hence, the work is completed in 25 days

*************************************

(7) A machine costs 5,00,000rs. If the value depreciates 15% in the first year, 13(1/2) in 2nd year 12% in the 3rd year and so on..What will be its value at the end of 10 years. When all the percentages will be applied to the original cost?

sol) Total depreciation:-

=> 15 + 13(1/2) + 12 + ----10 terms.

Here , a = 15

d1 = a2 - a1 = 13.5 - 15 = -1.5

d2 = a3 - a2 = 12 - 13.5 = -1.5

common diff same:- d1 = d2

Formulae :-

Sn = n [ 2a + (n-1) d]
        2

S10 = 10 [ 2(15) + (10-1)*(-1.5)]
         2

=> 5 (30 - 13.5)

=> 5 (16.5)

=> 82.5 %

Therefore, after ten year original cost :-

=> 100 - 82.5

=> 17.5%

i.e 17.5% of 5,00,000

SSC MATHEMATICS

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Saturday, 31 March 2018

Chapter 6 ) - Progressions

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