Exercise 5.3

1. Find the sum of the following APs:

(i)

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

Solution:

2, 7, 12,... , to 10 terms.

Given,

First term, a =

Common Difference, d = 7 - 2 =

Number of Terms, n =

We know that sum up to nth term of AP is given by formula Sn = n2 [2a + (n - 1) d]

S10 = 102 [2 × + ( - 1) ]

= 5 [4 + × 5]

= 5[4 + ]

= 5 ×

The sum of 10 terms is 245.

(ii)

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

Solution:

Given,

First term, a =

Common Difference, d = (- 33) - (- 37) =

Number of Terms, n =

We know that sum up to nth term of AP is given by formula Sn = n2 [2a + (n - 1) d]

S12 = 122 [2 (- ) + (12 - 1) 4]

= [- + × 4]

= 6 [- 74 + ]

= 6 × (- )

The sum of 12 terms is - 180.

(iii)

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

Solution:

Given,

First term, a =

Common Difference, d = 1.7 - 0.6 =

Number of Terms, n =

We know that sum up to nth term of AP is given by formula Sn = n2 [2a + (n - 1) d]

S100 = 1002 [2 × + (100 - 1) ]

= 50 [1.2 + × 1.1]

= 50 [1.2 + ]

= 50 [ ]

The sum of 100 terms is 5505.

(iv)

(iv) 115 , 112 , 110 ,...., to 11 terms

Solution:

Given,

First term, a =

Common Difference, d = 112 - 115 =

Number of Terms, n =

We know that sum up to nth term of AP is given by formula Sn = n2 [2a + (n - 1) d]

S11 = 112 [2 × 115 + ( - 1) × 160]

= 112 [215 + 16]

= 112 [ (4 + 5)30]

= 112 ×

2. Find the sums given below :

(i)

(i) 7 + 10 12 + 14 +... + 84

Solution:

Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.

Given,

First-term, a =

Common Difference, d = 10 12 - 7 = 212 - 7 =

Last term, l =

To find the total number of terms, we will use the formula an + (n - 1) d.

l = an = a + (n - 1) d

= 7 + (n - 1) 72

= (n - 1) 72

= n - 1

n =

Hence, n = 23

Sn = n2 [a + l]

S23 = 232 [7 + ]

= (232) ×

= 20932

= 12

(ii)

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

Solution:

Given,

First term, a =

Common Difference d= 32 - 34 =

Last term, l =

To find the total number of terms, we will use the formula an + (n - 1) d.

l = a + (n - 1) d

= + (n - 1) (- 2)

n - 1 =

n =

Hence, n = 13

Sn = n2 [a + l]

S13 = 132 ( + 10)

= (132) ×

= 13 ×

(iii)

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

Solution:

Given,

First-term, a=

Common Difference, d = (- 8) - (- 5) = - 8 + 5 =

Last term, l =

To find the total number of terms, we will use the formula an + (n - 1) d.

l = a + (n - 1) d

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

n - 1 =

n = + 1

n =

Hence, n = 76

Sn = n2 [a + l]

= 762 [(- 5) + ]

= 38 ×

In an AP:

(i)

(i)Given a = 5, d = 3, an = 50 , find n and Sn

Solution:

Sn = n2 [2a + (n - 1) d],

an= a + (n - 1)d

Here, a is the first term, d is the common difference, n is the number of terms and l is the last term.

Given,

First term, a =

Common difference, d =

nth term, l = an =

an= a + (n - 1)d

= 5 + (n - 1)3

= (n - 1)3

= n - 1

n =

Sustitute the value of n, to find sum of the n terms.

Sn = n2 [a + l]

S16 = 162 [ 5 + ]

= 8 ×

(ii)

(ii)Given a = 7, a13 = 35 , find d and S13

Solution:

Given,

First term, a =

13th term, l = a13 =

lan = a + (n - 1)d__

a13 = a + ( - 1) d

35 = 7 + d

35 - = 12d

d = 2812

d =

Sn = n2 [a + l]

S13 = 132 [7 + ]

= 132 ×

= 13 ×

(iii)

(iii)Given a12 = 37, d = 3, find a and S12.

Solution:

Given,

12th term, a12 =

Common Difference, d =

an = a + (n - 1)d

a12 = a + ( - 1) 3

37 = a +

a =

Substitute the value of a to find the sum of n terms.

Sn = n2 [a + l]

S12 = 122 [4 + ]

= 6 ×

S12 =

(iv)

(iv)Given a3 = 15, S10 = 125, find d and a10.

Solution:

Given,

3rd term, a3 =

Sum up to ten terms, S10 = .

an = a + (n - 1)d

a3 = a + ( - 1) d

15 = a + d - - - - - Equation (i)

Sn = n2 [2a + (n - 1) d]

S10 = 102 [2a + ( - 1) d]

125 = [2a + d]

= 2a + 9d - - - - - Equation (ii)

On multiplying equation (i) by 2, we obtain

= 2a + d - - - - - Equation (iii)

On subtracting equation (iii) from Equation (ii), we obtain

- 5 = 5d

d =

From equation (i),

15 = a + 2 (- 1)

15 = a -

a =

a10 = a + ( - 1) d

a10 = 17 + (- 1)

a10 = 17 -

a10 =

(v)

(v)Given d = 5, S9 = 75 , find a and a9.

Solution:

Given,

Common difference, d =

Sum up to nine terms, S9 =

Sn = n2 [2a + (n - 1) d]

= 92 [2a + ( - 1) 5]

75 = 92 (2a + )

= 3(a + 20)

25 = 3a +

a =

We know that nth term of the AP is given by formula an= a + (n - 1)d

a9 = a + ( - 1) × 5

= −353 + × 5

= −353 +

= (- 35 + 120)3

In an AP:

(vi)

(vi)Given a = 2, d = 8, Sn = 90 , find n and an.

Solution:

Given,

First term,a =

Common difference, d =

Sum up to nth terms, Sn =

Sn = n2 [2a + (n - 1) d]

90 = n2 [4 + (n - 1) ]

90 = n [2 + (n - 1)]

90 = n [2 + - 4]

90 = n [4n - ]

90 = 4n2 -

4n2 - 2n - = 0

4n2 - 20n + n - 90 = 0

4n (n - ) + 18(n - ) = 0

(n - 5)(4n + 18) =

Either (n - 5) = or (4n + 18) = 0

n = or n = −92

However, n can neither be negative nor fractional.

Therefore, n = 5

an = a + (n - 1) d

a5 = 2 + ( - 1)8

a5 = 2 + × 8

a5 = 2 +

a5 =

(vii)

(vii)Given a = 8, an = 62, Sn = 210 , find n and d.

Solution:

Given,

Given,

First term, a =

nth term, l = an =

Sum up to nth terms, Sn =

Sn = n2 [2a + (n - 1) d]

210 = n2 [8 + ]

210 = n ×

n =

We know that nth term of the AP is given by formula an= a + (n - 1)d

62 = 8 + ( - 1) d

62 - 8 = d

= 5d

d =

(viii)

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

Solution:

Given,

Common difference d =

nth term, l = an=

Sum up to nth terms, Sn =

We know that nth term of the AP is given by formula an= a + (n - 1)d

4 = a + (n - 1)2

4 = a + - 2

a = - 2n .... (1)

Sn = n2 [a + l]

- = n2 [6 - 2n + 4] [from(1)]

- 14 = n ( - n)

- 14 = 5n - n2

n2 - 5n - = 0

n2 - n + 2n -14 = 0

n (n - 7) + (n - 7) = 0

(n - 7)(n + 2) =

Either n - 7 = or n + 2 = 0

n = 7 or n =

However, n can neither be negative nor fractional.

Therefore, n =

From equation (1), we obtain

a = 6 - 2n

a = 6 - 2 ×

a = 6 -

a =

(ix)

(ix)Given a = 3, n = 8, S = 192 , find d.

Solution:

Given,

First term, a =

Number of terms, n =

Sum up to nth terms, Sn=

Sn= n2 [2a + (n - 1) d]

192 = 82 [2 × + (8 - 1) d]

192 = [6 + d]

= 6 + 7d

= 7d

d =

(x)

(x)Given l = 28, S = 144 and there are total 9 terms. Find a.

Solution:

Given,

Last term, l = an =

Number of terms, n =

Sum up to nth terms, Sn =

Sn = n2 [a + l]

144 = 92 (a + )

= a + 28

a =

4. How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Solution:

Sn = n2 [2a + (n - 1) d]

Here, a is the first term, d is the common difference, n is the number of terms and l is the last term.

Given,

First term, a =

Common difference, d = 17 - 9 = 25 - 17 = ... =

Sum up to nth terms, Sn =

We know that sum of n terms of an AP Sn = n2 [2a + (n - 1) d]

= n2 [2 × + (n - 1) 8]

636 = n2 ( + 8n - 8)

636 = n2 ( + 8n)

636 = n(5 + )

636 = + 4n2

4n2 + 5n - = 0

4n2 + n - n - 636 = 0

n (4n + 53) - (4n + 53) = 0

(4n + 53)(n - 12) =

Either 4n + 53 = or n - 12 = 0

n = −534 or n =

n cannot be −534 because the number of terms can neither be negative nor fractional, therefore, n = 12

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution:

Sn = n2 [2a + (n - 1) d]

Here, a is the first term, d is a common difference and n is the number of terms.

Given,

First term, a =

Last term, l =

Sum of n terms, Sn =

We know that sum of n terms of AP is given by the formula Sn = n2 [a + l]

= n2 (5 + )

400 = n2 ×

n =

By using the formula an = a + (n -1) d, we will find the common difference where an = l.

l = an = a + (n -1) d

45 = + (16 -1) d

= d

d = 4015

d =

6. 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?

Solution:

Sn = n2 [2a + (n - 1) d]

Here, a is the first term, d is a common difference and n is the number of terms.

Given,

First term, a =

Last term, l =

Common difference, d =

We know that nth term of an AP, l = an = a + (n - 1)d

= + (n - 1) 9

= (n - 1) 9

(n - 1) =

n =

Sum of n terms of AP, Sn = n2 [a + l]

S38 = 382 ( + 350)

= ×

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Solution:

Sn = n2 [2a + (n - 1) d]

Here, a is the first term, d is a common difference and n is the number of terms.

Given,

22nd term, l = a22 = 149

Common difference, d =

We know that nth term of AP, an = a + (n - 1)d

a22 = a + ( - 1) d

= a + × 7

149 = a +

a =

The sum of 22 terms can be found as follows.

Sn = n2 [a + l]

= 222 ( + 149)

= 11 ×

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

Solution:

Sn = n2 [2a + (n - 1) d]

Here, a is the first term, d is a common difference and n is the number of terms.

Given,

2nd term, a2 =

3rd term, a3=

Common difference, d = a3 - a2 = 18 - 14 =

We know that nth term of AP, an = a + (n - 1)d

a2 = a +

= a + 4

a =

Sum of n terms of AP is given by Sn = n2 [2a + (n - 1) d]

S51 = 512 [2 × + (51 - 1) 4]

= 512 [20 + × 4]

= 512 ×

= 51 ×

9. 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.

Solution:

Sn = n2 [2a + (n - 1) d]

Here, a is the first term, d is a common difference and n is the number of terms.

Given,

Sum of first 7 terms, S7 =

Sum of first 17 terms, S17 =

We know that sum of n terms of AP is Sn = n2 [2a + (n - 1) d]

S7 = 72 [2a + ( - 1)d]

49 = 72 [2a + ]

a + 3d = ... (i)

S17 = 172 [2a + ( - 1) d]

289 = 172 [2a + d]

a + 8d = ... (ii)

Subtracting equation (i) from equation (ii),

a + 8d - (a + 3d) = 17 -

5d =

d =

From equation (i),

7 = a + 3 ×

7 = a +

a =

Sn = n2 [2a + (n - 1) d]

= n2 [2 × + (n - 1) 2]

= n2 [2 + - 2]

= n2 ×

10. Show that a1, a2, . . ., an, . . . form an AP where an is defined as below :

(i)

(i) an = 3 + 4n

Solution:

Given,

nth term, an = 3 + 4n

a1 = 3 + 4 × =

a2 = 3 + 4 × = 3 + =

a3 = 3 + 4 × = 3 + =

a4 = 3 + 4 × = 3 + =

It can be observed that

a2 - a1 = 11 - 7 =

a3 - a2 = 15 - 11 =

a4 - a3 = 19 - 15 =

So, the difference of an and an1 is constant.

Therefore, this is an AP with common difference as 4 and first term as .

The sum of n terms of AP is given by thre formula Sn = n2 [2a + (n - 1) d]

Sum of 15 terms,

S15 = 152 [2 × 7 + ( - 1) 4]

= 152 [14 + × 4]

= 152 ×

= 15 ×

(ii)

(ii) an = 9 - 5n

Solution:

Given,

nth term is an = 9 - 5n

a1 = 9 - 5 × 1 = 9 - =

a2 = 9 - 5 × 2 = 9 - =

a3 = 9 - 5 × 3 = 9 - =

a4 = 9 - 5 × 4 = 9 - =

It can be observed that

a2 -a1 = (-1 ) - 4 =

a3 - a2 = (-6) -(-1) =

a4 - a3 = (- 11) - (- 6) =

So, the difference ofan and an1 is constant.

Therefore, this is an A.P. with common difference - 5 and first term as .

The sum of n terms of AP is given by thre formula Sn = n2 [2a + (n - 1) d]

S15 = 152 [2 × + (15 - 1)(- 5)]

= 152 [8 + (- 5)]

= 152 [8 - ]

= 152 × ()

11. If the sum of the first n terms of an AP is 4n - n2, what is the first term (that is S1)? What is the sum of the first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Solution:

Sn = n2 [2a + (n - 1) d],

an = a + (n - 1)d

Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.

Sum of first n terms, Sn = 4n - n2

Therefore,

Sum of first term, S1 = 4 × 1 - 12 = 4 - =

Sum of first two terms, S2 = 4 × 2 - 22= 8 - =

Sum of first three terms, S3 = 4 × 3 - 32 = 12 - =

Second term,a2 =S2 - S1 = 4 - 3 =

Third term, a3 = S3 - S2 = 3 - 4 =

Tenth term, a10 = S10 - S9

= (4 × 10 - 102) - (4 × 9 - 92)

= (40 - ) - (36 - )

= -60 +

nth term, an = Sn - Sn1

= [4n - n2] - [4 (n - 1) - n−12 ]

= 4n - n2 - + 4 + n−12

= 4 - n2 + n2 - 2n +

= 5 -

12. Find the sum of the first 40 positive integers divisible by 6.

Solution:

Sn = n2 [2a + (n - 1) d],

an = a + (n - 1)d

Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.

The positive integers that are divisible by 6 are , , , , ...

It can be observed that these numbers are forming an AP.

Hence,

First term, a =

Common difference, d =

Number of terms, n =

As we know that sum of n terms is given by the formula Sn = n2 [2a + (n - 1) d],

S40 = 402 [2 × + (40 - 1)6]

= 20 [12 + × 6]

= 20 [12 + ]

= 20 ×

13. Find the sum of the first 15 multiples of 8.

Solution:

Here, a is the first term, d is a common difference and n is the number of terms.

The multiples of 8 are 8, , , , ...

These numbers are in an A.P.

Hence,

First term, a =

Common difference, d =

Number of terms, n =

As we know that sum of n terms is given by the formula Sn = n2 [2a + (n - 1) d],

S15 = 152 [2 × 8 + ( - 1)8]

= 152 [16 + × 8]

= 152 [16 + ]

= 152 ×

= 15 ×

14. Find the sum of the odd numbers between 0 and 50.

Solution:

Sn = n2 [2a + (n - 1) d],

Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.

The odd numbers lying between 0 and 50 are 1, , , , ...

Therefore, it can be observed that these odd numbers are in an A.P.

Hence,

First term, a =

Common difference, d =

Last term, l=

We know that nth term of AP, an = a + (n - 1)d

= 1 + (n - 1) 2

= 2(n - 1)

n - 1 =

n =

We know that sum of n terms of AP,

Sn = n2 [a + l]

S25 = 252 (1 + )

= 252 ×

= 25 ×

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty if he has delayed the work by 30 days?

Solution:

Sn = n2 [2a + (n - 1) d],

Here, a is the first term, d is a common difference and n is the number of terms.

Penalty for 1st day = ₹

Penalty for 2nd day = ₹

Penalty for 3rd day = ₹

By observation, penalties are in A.P. having the first term as 200 and common difference as 50 and number of terms as .

a = , d = , n =

Penalty that has to be paid if he has delayed the work by 30 days will be calculated as follows.

Sn = n2 [2a + (n - 1) d]

S30 = 302 [2 × 200 + ( - 1) 50]

= 15 [400 + ]

= 15 ×

Therefore, the contractor has to pay ₹ 27750 as a penalty.

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

Solution:

7 cash prizes are given, and each prize is ₹ less than its preceding prize.

The sum of the first n terms of an AP is given by Sn = n2 [2a + (n - 1) d],

Here, a is the first term, d is a common difference and n is the number of terms.

Let the cost of the 1st prize be x

The cost of 2nd prize = x -

The cost of 3rd prize = x -

Thus, the prizes are x, (x - 20), (x - 40),...

By observation, costs of these prizes are in an A.P., having common difference as - 20 and first term as x.

a = x, d =

Given that, S7 =

By using the formula Sn = n2 [2a + (n - 1) d]

72 [2x + ( - 1) d] = 700

7 [2x + () × (- 20)] = 700 ×

[2x + (6) × (- 20)] = 14007

[2x + (6) × (- 20)] =

x + 3 × (- 20) =

x - = 100

x =

Therefore, the value of each of the prizes was Rs. 160, Rs. 140, Rs. 120, Rs. 100, Rs. 80, Rs. 60, and Rs. 40.

17. 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 the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution:

Sn = n2 [2a + (n - 1) d],

an = a + (n - 1)d

Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.

-	Class I	Class II	Class III	-----	Class XII
Section A	1	2	3	----	12
Section B	1	2	3	----	12
Section C	1	2	3	----	12
Total				----

It can be observed that the number of trees planted by the students are in an AP.

3, , , , ,..., 36

First term, a =

Common difference, d = 6 - 3 =

Number of terms, n =

We know that sum of n terms of AP, Sn = n2 [2a + (n - 1) d]

S12 = 122 [2 × 3 + ( - 1) × 3]

= 6 [6 + × 3]

= 6 [6 + ]

= 6 ×

Therefore, 234 trees will be planted by the students.

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . .as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? [Take π = 227]

Solution:

Sn = n2 [2a + (n - 1) d]

Here, a is the first term, d is the common difference and n is the number of terms.

The Semiperimeter of a circle, l = πr

Given that the radii of the spiral follows a pattern of 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,...

l1 = π × (0.5 cm) = π cm [Since, radius = 0.5 cm]

l2 = π × (1 cm) = cm [Since, radius = 1 cm]

l3 = π × (1.5 cm) = π cm [Since, radius = 1.5 cm]

Therefore, l1, l2, l3, ... the lengths of the semi-circles are in an A.P

So,the AP is 0.5π, π, 1.5π, 2π, ...

a = π

d = π - 0.5π = π

We know that the sum of n terms of an A.P. is given by Sn = n2 [2a + (n - 1) d]

We need to calculate the total length of such a spiral made up of thirteen consecutive semicircles

Thus, n =

S13 = 132 [2 × (0.5π) + ( - 1)(0.5π) ]

= 132 [ π + π]

= 132 × π

= 132 × 7 × 227 (By taking π = 227)

Therefore, the length of such a spiral made up of thirteen consecutive semi-circles will be 143 cm.

19. 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 and so on (see Fig). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:

Sn = n2 [2a + (n - 1) d]

Here, a is the first term, d is the common difference and n is the number of terms.

It can be observed that the number of logs in rows are forming an A.P. 20, 19, 18, ...

First terms, a =

Common difference, d = 19 - 20 =

Sum of the n terms, Sn =

We know that the sum of n terms of an A.P. is given by Sn = n2 [2a + (n - 1) d]

200 = n2 [2 × + (n - 1)(- 1)]

400 = n [40 - + 1]

400 = n ( - n)

400 = - n2

n2 - 41n + = 0

n2 - n - n + 400 = 0

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

(n - 16)(n - 25) =

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

∴ n = or n =

The number of logs in nth row will be,

an = a + (n - 1) d

a16 = 20 + ( -1) × (- 1)

a16 = 20 -

a16 =

Similarly,

a25 = 20 + ( - 1) × (- 1)

a25 = 20 -

a25 =

Clearly, the number of logs in the 16th row is 5. However, the number of logs in the 25th row is negative 4, which is not possible.

Therefore, 200 logs can be placed in 16 rows. The number of logs in the top (16th) row is 5.

Arithmetic Progressions

Glossary

Exercise 5.3

(i)

(ii)

(iii)

(iv)

(i)

(ii)

(iii)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(i)

(ii)

Sign in to Innings2

Arithmetic Progressions

Reset Progress

Glossary

Exercise 5.3

(i)

(ii)

(iii)

(iv)

(i)

(ii)

(iii)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(i)

(ii)