Arithmetic Progression — Exercise 10(C)

The sum of 41 terms of an A.P. with middle term 40 is :

1. 820

2. 1640

3. 2460

4. none of these

Answer

No. of terms = 41

Middle term = $\dfrac{41 + 1}{2} = \dfrac{42}{2}$ = 21st term

Given,

Middle term = 40

⇒ a₂₁ = 40

⇒ a + (21 - 1)d = 40

⇒ a + 20d = 40 ............(1)

By formula,

Sum of A.P. = $\dfrac{n}{2}$[2a + (n - 1)d]

= $\dfrac{41}{2}[2a + (41- 1)d]$

= $\dfrac{41}{2}[2a + 40d]$

= $\dfrac{41}{2} \times 2 \times [a + 20d]$

= 41 × 40 ..........[From (1)]

= 1640.

Hence, Option 2 is the correct option.

The sum of all two digit numbers is :

1. 9810

2. 9045

3. 4509

4. 4905

Answer

Two digit numbers : 10, 11, ......., 99.

The above list is an A.P. with first term (a) = 10 and common difference (d) = 11 - 10 = 1.

Last term = 99.

Let n^th term be the last term.

∴ a_n = a + (n - 1)d

⇒ 99 = 10 + (n - 1)1

⇒ 99 = 10 + n - 1

⇒ 99 = n + 9

⇒ n = 99 - 9 = 90.

By formula,

Sum of n terms = $\dfrac{n}{2}[2a + (n - 1)d]$

Sum of the above A.P.

$= \dfrac{90}{2}[2 \times 10 + (90 - 1)(1)] \\[1em] = 45[20 + 89] \\[1em] = 45 \times 109 \\[1em] = 4905.$

Hence, Option 4 is the correct option.

The sum of A.P. 4, 7, 10, 13, ........ upto 20 terms is :

1. 650

2. 10 × 27

3. 510

4. 1300

Answer

In A.P. 4, 7, 10, 13, ........ upto 20 terms

First term (a) = 4

Common difference (d) = 7 - 4 = 3

By formula,

Sum of n terms = $\dfrac{n}{2}[2a + (n - 1)d]$

Sum of above A.P. (upto 20 terms)

$= \dfrac{20}{2}[2 \times 4 + (20 - 1) \times 3] \\[1em] = 10 \times [8 + 19 \times 3] \\[1em] = 10 \times [8 + 57] \\[1em] = 10 \times 65 \\[1em] = 650.$

Hence, Option 1 is the correct option.

The sum of 40 terms of the A.P. 7 + 10 + 13 + 16 + ..... is :

1. 5240

2. 2620

3. 1310

4. 2680

Answer

In the A.P., 7 + 10 + 13 + 16 + .........

First term (a) = 7

Common difference (d) = 10 - 7 = 3.

By formula,

Sum of n terms = $\dfrac{n}{2}[2a + (n - 1)d]$

Sum of above A.P. (upto 20 terms)

$= \dfrac{40}{2}[2 \times 7 + (40 - 1) \times 3] \\[1em] = 20 \times [14 + 39 \times 3] \\[1em] = 20 \times [14 + 117] \\[1em] = 20 \times 131 \\[1em] = 2620.$

Hence, Option 2 is the correct option.

The n^th term of an A.P. is 6n + 4. The sum of its first two terms is :

1. 16

2. 20

3. 26

4. none of these

Answer

Given,

n^th term = 6n + 4

First term = 6(1) + 4 = 10

Second term = 6(2) + 4 = 12 + 4 = 16.

Sum of first two terms = 10 + 16 = 26.

Hence, Option 3 is the correct option.

How many terms of the A.P. :

24, 21, 18, ........... must be taken so that their sum is 78 ?

Answer

In above A.P. a = 24 and d = -3.

We know that,

S = $\dfrac{n}{2}(2a + (n - 1)d)$

Let sum of n terms be 78.

$\Rightarrow 78 = \dfrac{n}{2}(2 \times 24 + (n - 1)(-3)) \\[1em] \Rightarrow 78 = \dfrac{n}{2}(48 - 3n + 3) \\[1em] \Rightarrow 78 \times 2 = n(51 - 3n) \\[1em] \Rightarrow 156 = 51n - 3n^2 \\[1em] \Rightarrow 3n^2 - 51n + 156 = 0 \\[1em] \Rightarrow 3(n^2 - 17n + 52) = 0 \\[1em] \Rightarrow n^2 - 17n + 52 = 0 \\[1em] \Rightarrow n^2 - 13n - 4n + 52 = 0 \\[1em] \Rightarrow n(n - 13) - 4(n - 13) = 0 \\[1em] \Rightarrow (n - 4)(n - 13) = 0 \\[1em] \Rightarrow n - 4 = 0 \text{ or } n - 13 = 0 \\[1em] \Rightarrow n = 4, 13.$

Hence, no. of terms = 4 or 13.

Find the sum of 28 terms of an A.P. whose n^th term is 8n - 5.

Answer

Given,

a_n = 8n - 5

So,

a₁ = 8(1) - 5 = 3,

a₂₈ = 8(28) - 5 = 219.

$S = \dfrac{n}{2}(a + l) \\[1em] = \dfrac{28}{2}(3 + 219) \\[1em] = 14 \times 222 \\[1em] = 3108.$

Hence, sum = 3108.

Find the sum of all odd natural numbers less than 50.

Answer

Odd numbers less than 50 = 1, 3, 5, ........., 49.

The above series is an A.P. with a = 1, d = 2 and l = 49.

Let there be n terms in series,

⇒ a_n = 49

⇒ a + (n - 1)d = 49

⇒ 1 + 2(n - 1) = 49

⇒ 1 + 2n - 2 = 49

⇒ 2n - 1 = 49

⇒ 2n = 50

⇒ n = 25.

We know that,

$S = \dfrac{n}{2}(a + l) \\[1em] = \dfrac{25}{2}(1 + 49) \\[1em] = \dfrac{25}{2} \times 50 \\[1em] = 625.$

Hence, sum of odd natural numbers less than 50 = 625.

Find the sum of first 12 natural numbers each of which is a multiple of 7.

Answer

First 12 natural numbers that are a multiple of 7 are,

7, 14, 21, ..........., 12^th term.

The above sequence is an A.P. with a = 7 and common difference = 7.

$S = \dfrac{n}{2}(2a + (n - 1)d) \\[1em] = \dfrac{12}{2}(2 \times 7 + (12 - 1) \times 7) \\[1em] = 6(14 + 77) \\[1em] = 6 \times 91 \\[1em] = 546.$

Hence, sum of first 12 natural numbers that are divisible by 7 is 546.

Find the sum of first 51 terms of an A.P. whose 2^nd and 3^rd terms are 14 and 18 respectively.

Answer

Given,

⇒ a₂ = a + (2 - 1)d

⇒ 14 = a + d

⇒ a = 14 - d .......(i)

Also,

⇒ a₃ = a + (3 - 1)d

⇒ 18 = a + 2d

⇒ a = 18 - 2d .......(ii)

From (i) and (ii) we get,

⇒ 14 - d = 18 - 2d

⇒ -d + 2d = 18 - 14

⇒ d = 4.

Substituting value of d in (i) we get,

⇒ a = 14 - 4 = 10.

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{51}{2}[2 \times 10 + (51 - 1) \times 4] \\[1em] = \dfrac{51}{2} \times (20 + 200) \\[1em] = \dfrac{51}{2} \times 220 \\[1em] = 51 \times 110 \\[1em] = 5610.$

Hence, sum of first 51 terms = 5610.

The sum of first 7 terms of an A.P. is 49 and that of first 17 terms of it is 289. Find the sum of first n terms.

Answer

Let the first term of A.P. be a and common difference = d.

Given, sum of first 7 terms of an A.P. is 49,

$\Rightarrow S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 49 = \dfrac{7}{2}[2 \times a + (7 - 1)d] \\[1em] \Rightarrow 49 = \dfrac{7}{2}[2a + 6d] \\[1em] \Rightarrow 49 = 7(a + 3d) \\[1em] \Rightarrow a + 3d = 7 ........(i)$

Given, sum of first 17 terms of an A.P. is 289,

$\Rightarrow S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 289 = \dfrac{17}{2}[2 \times a + (17 - 1)d] \\[1em] \Rightarrow 289 = \dfrac{17}{2}[2a + 16d] \\[1em] \Rightarrow 289 = 17(a + 8d) \\[1em] \Rightarrow a + 8d = 17 ........(ii)$

Subtracting (i) from (ii) we get,

⇒ a + 8d - (a + 3d) = 17 - 7

⇒ a - a + 8d - 3d = 10

⇒ 5d = 10

⇒ d = 2.

Substituting value of d in (i) we get,

⇒ a + 3(2) = 7

⇒ a + 6 = 7

⇒ a = 1.

$\text{Sum of n terms } = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{n}{2}[2(1) + (n - 1)2] \\[1em] = \dfrac{n}{2}[2 + 2n - 2] \\[1em] = \dfrac{n}{2} \times 2n \\[1em] = n^2.$

Hence, sum of n terms = n².

The first term of an A.P. is 5, the last term is 45 and the sum of its term is 1000. Find the number of terms and the common difference of the A.P.

Answer

Given, a = 5, l = 45 and S = 1000.

$\Rightarrow S = \dfrac{n}{2}(a + l) \\[1em] \Rightarrow 1000 = \dfrac{n}{2}(5 + 45) \\[1em] \Rightarrow 1000 = \dfrac{n}{2} \times 50 \\[1em] \Rightarrow n = \dfrac{1000 \times 2}{50} \\[1em] \Rightarrow n = 40.$

Given, l = a₄₀ = 45

⇒ a + (40 - 1)d = 45
⇒ 5 + 39d = 45
⇒ 39d = 40
⇒ d = $\dfrac{40}{39}$.

Hence, n = 40 and d = $\dfrac{40}{39}$.

Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.

Answer

$\dfrac{250}{9} = 27\dfrac{7}{9} \text{ and } \dfrac{1000}{9} = 111\dfrac{1}{9}$.

The numbers which are divisible by 9 between 250 and 1000 are,

= 28 × 9, 29 × 9, 30 × 9, ............, 111 × 9.

= 252, 261, 270, .........., 999.

The above sequence is an A.P. with common difference = 9 and first term = 252 and last term = 999.

Let n be no. of terms,

∴ a_n = a + (n - 1)d

⇒ 999 = 252 + (n - 1)9

⇒ 999 = 252 + 9n - 9

⇒ 999 = 9n + 243

⇒ 999 - 243 = 9n

⇒ 9n = 756

⇒ n = 84.

$S = \dfrac{n}{2}(a + l) \\[1em] = \dfrac{84}{2} \times (252 + 999) \\[1em] = 42 \times 1251 \\[1em] = 52542.$

Hence, sum = 52542.

The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?

Answer

Let n be no. of terms,

∴ a_n = a + (n - 1)d

⇒ 700 = 34 + (n - 1)18

⇒ 700 = 34 + 18n - 18

⇒ 700 = 18n + 16

⇒ 700 - 16 = 18n

⇒ 18n = 684

⇒ n = 38.

$S = \dfrac{n}{2}(a + l) \\[1em] = \dfrac{38}{2} \times (34 + 700) \\[1em] = 19 \times 734 \\[1em] = 13946.$

Hence, no. of terms = 38 and sum = 13946.

In an A.P. the first term is 25, n^th term is -17 and the sum of n terms is 132. Find n and the common difference.

Answer

Given,

a = 25, a_n = -17 and S = 132

⇒ a + (n - 1)d = -17

⇒ 25 + (n - 1)d = -17

⇒ (n - 1)d = -42 ........(i)

We know that,

$\Rightarrow S = \dfrac{n}{2}(2a + (n - 1)d) \\[1em] \Rightarrow 132 = \dfrac{n}{2}(2 \times 25 + (-42)) \\[1em] \Rightarrow 132 = \dfrac{n}{2}(50 - 42) \\[1em] 132 = \dfrac{8n}{2} \\[1em] 4n = 132 \\[1em] n = 33.$

Substituting value of n in (i),

⇒ (33 - 1)d = -42

⇒ 32d = -42

⇒ d = -$\dfrac{42}{32} = -\dfrac{21}{16}$

Hence, n = 33 and d = $-\dfrac{21}{16}$.

If 18, a and (b - 3) are in A.P., then find the value of (2a - b).

Answer

Given,

18, a, (b - 3) are in A.P.

In an A.P., consecutive differences are equal.

⇒ a - 18 = (b - 3) - a

⇒ a - 18 = b - 3 - a

⇒ a + a - 18 + 3 = b

⇒ 2a - 15 = b

⇒ 2a - b = 15

Hence, the value of 2a - b = 15.

Find the A.P. whose 4^th term is 9 and the sum of its 6^th term and 13^th term is 40.

Answer

Given,

a₄ = 9

⇒ a + (4 - 1)d = 9

⇒ a + 3d = 9

⇒ a = 9 - 3d ........(1)

Given,

Sum of a₆ + a₁₃ = 40

⇒ a + (6 - 1)d + a + (13 - 1)d = 40

⇒ a + 5d + a + 12d = 40

⇒ 2a + 17d = 40 ......(2)

Substituting value of a from equation (1) in equation (2), we get :

⇒ 2(9 − 3d) + 17d = 40

⇒ 18 − 6d + 17d = 40

⇒ 18 + 11d = 40

⇒ 11d = 22

⇒ d = 2.

Substituting the value of d in equation (1), we get :

⇒ a = 9 - 3(2)

⇒ a = 9 - 6

⇒ a = 3.

So the A.P. is,

3, 5, 7, 9, 11, ....

Hence, A.P is 3, 5, 7, 9, 11, .....

The sum of n natural numbers is 5n² + 4n. Find its 8^th term.

Answer

Here T stands for term.

S_n = 5n² + 4n

S₁ = Sum of first natural number = T₁

= 5(1)² + 4(1) = 5 + 4 = 9.

S₂ = 5(2)² + 4(2) = 20 + 8 = 28.

T₂ = S₂ - S₁ = 28 - 9 = 19.

We know that,

⇒ T₂ = a + d

⇒ 19 = 9 + d

⇒ d = 10.

T₈ = a + (8 - 1)d = 9 + 7(10) = 79.

Hence, 8^th term = 79.

The fourth term of an A.P. is 11 and the eight term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.

Answer

Let the first term of an A.P. be a and common difference be d.

Given,

⇒ a₄ = 11

⇒ a + (4 - 1)d = 11

⇒ a + 3d = 11 .......(i)

Also,

⇒ a₈ = 2a₄ + 5

⇒ a + (8 - 1)d = 2[a + (4 - 1)d] + 5

⇒ a + 7d = 2a + 6d + 5

⇒ a - 2a + 7d - 6d = 5

⇒ -a + d = 5 ........(ii)

Adding (i) and (ii) we get,

⇒ a + 3d + (-a + d) = 11 + 5

⇒ 4d = 16

⇒ d = 4.

Substituting value of d in (i) we get,

⇒ a + 3(4) = 11

⇒ a + 12 = 11

⇒ a = -1.

A.P. = a, (a + d), (a + 2d), ..........

= -1, 3, 7, ...........

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{50}{2}[2 \times (-1) + (50 - 1) \times 4] \\[1em] = 25[-2 + 196] \\[1em] = 25 \times 194 \\[1em] = 4850.$

Hence, A.P. = -1, 3, 7, ........... and sum of first 50 terms = 4850.