Arithmetic & Geometric Progression — Exercise 9.3

Find the sum of the following A.P.s :

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

(ii) $\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10}, ...$ to 11 terms

Answer

(i) Here, a = 2, d = 7 - 2 = 5 and n = 10.

We know that Sum = $\dfrac{n}{2}[2a + (n - 1)d]$

$\therefore \text{ Sum } = \dfrac{10}{2}[2 \times 2 + (10 - 1)5] \\[1em] = 5[4 + 9 \times 5] \\[1em] = 5 \times 49 \\[1em] = 245.$

Hence, the sum of the A.P. 2, 7, 12, ... upto 10 terms is 245.

(ii) Here, a = $\dfrac{1}{15}, d = \dfrac{1}{12} - \dfrac{1}{15} = \dfrac{5 - 4}{60} = \dfrac{1}{60}$ and n = 11.

We know that Sum = $\dfrac{n}{2}[2a + (n - 1)d]$

$\therefore \text{ Sum } = \dfrac{11}{2}[2 \times \dfrac{1}{15} + (11 - 1)\dfrac{1}{60}] \\[1em] = \dfrac{11}{2}\Big[\dfrac{2}{15} + \dfrac{10}{60}\Big] \\[1em] = \dfrac{11}{2} \times \Big[\dfrac{2 \times 4 + 10}{60}\Big] \\[1em] = \dfrac{11}{2} \times \dfrac{18}{60}$

Dividing numerator and denominator by 6,

$= \dfrac{198}{120} \\[1em] = \dfrac{33}{20}. \\[1em]$

Hence, the sum of the A.P. $\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10}, ...$ upto 11 terms is $\dfrac{33}{20}$.

Find the sums given below:

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

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

Answer

(i) The given numbers form an A.P. with a = 34 and d = 32 - 34 = -2.

Let nth term be 10,
as, a_n = a + (n - 1)d

∴ 10 = 34 + (n - 1)(-2)
⇒ 10 = 34 - 2n + 2
⇒ 10 = 36 - 2n
⇒ 2n = 36 - 10
⇒ 2n = 26
⇒ n = 13.

$\text{Sum } = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \therefore \text{ Sum } = \dfrac{13}{2}[2 \times 34 + (13 - 1)(-2)] \\[1em] = \dfrac{13}{2}[68 + 12(-2)] \\[1em] = \dfrac{13}{2}[68 - 24] \\[1em] = \dfrac{13}{2} \times 44 \\[1em] = 13 \times 22 \\[1em] = 286.$

Hence, the sum of the series 34 + 32 + 30 + .... + 10 is 286.

(ii) The given numbers form an A.P. with a = -5 and d = -8 - (-5) = -3.

Let nth term be -230, then as, a_n = a + (n - 1)d

∴ -230 = -5 + (n - 1)(-3)
⇒ -230 = -5 -3n + 3
⇒ -230 = -2 - 3n
⇒ -230 + 2 = -3n
⇒ -228 = -3n
⇒ 3n = 228
⇒ n = 76.

$\text{Sum } = \dfrac{n}{2}[2a + (n - 1)d] \\[1 em] \therefore \text{Sum } = \dfrac{76}{2}[2 \times -5 + (76 - 1)(-3)] \\[1em] = 38[-10 + 75(-3)] \\[1em] = 38[-10 - 225] \\[1em] = 38 \times -235 \\[1em] = -8930.$

Hence, the sum of the series -5 + (-8) + (-11) + .... + (-230) is -8930.

In an A.P. (with usual notations):

(i) given a = 5, d = 3, a_n = 50, find n and S_n

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

(iii) given d = 5, S₉ = 75, find a and a₉

(iv) given a = 8, a_n = 62, S_n = 210, find n and d

(v) given a = 3, n = 8, S = 192, find d.

Answer

(i) a = 5, d = 3, a_n = 50.

By formula a_n = a + (n - 1)d

⇒ 50 = 5 + (n - 1)3
⇒ 50 = 5 + 3n - 3
⇒ 50 = 2 + 3n
⇒ 50 - 2 = 3n
⇒ 48 = 3n
⇒ n = 16.

By formula $S_n = \dfrac{n}{2}[2a + (n - 1)d]$

$S_n = \dfrac{16}{2}[2 \times 5 + (16 - 1)3] \\[1em] = 8[10 + 15 \times 3] \\[1em] = 8 \times 55 \\[1em] = 440.$

Hence, n = 16 and S_n = S₁₆ = 440.

(ii) a = 7, a₁₃ = 35

By formula a_n = a + (n - 1)d

⇒ 35 = 7 + (13 - 1)d
⇒ 35 = 7 + 12d
⇒ 35 - 7 = 12d
⇒ 28 = 12d
⇒ d = $\dfrac{28}{12}$ (Dividing by 4)

⇒ d = $\dfrac{7}{3}.$

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$S_{13} = \dfrac{13}{2}[2 \times 7 + (13 - 1)\Big(\dfrac{7}{3}\Big)] \\[1em] = \dfrac{13}{2}[14 + 12 \times \dfrac{7}{3}] \\[1em] = \dfrac{13}{2} \times [14 + 28] \\[1em] = \dfrac{13}{2} \times 42 \\[1em] = 13 \times 21 \\[1em] = 273.$

Hence, d = $\dfrac{7}{3}$ and S_n = S₁₃ = 273.

(iii) d = 5, S₉ = 75

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$S_9 = \dfrac{9}{2}[2 \times a + (9 - 1)5] \\[1em] \Rightarrow 75 = \dfrac{9}{2}[2a + 8 \times 5] \\[1em] \Rightarrow 75 = \dfrac{9}{2} \times [2a + 40] \\[1em] \Rightarrow 75 = 9(a + 20) \\[1em] \Rightarrow 75 = 9a + 180 \\[1em] \Rightarrow 9a = 75 - 180 \\[1em] \Rightarrow 9a = -105 \\[1em] \Rightarrow a = -\dfrac{105}{9} \\[1em] \Rightarrow a = -\dfrac{35}{3}$

By formula a_n = a + (n - 1)d,

$\Rightarrow a_9 = -\dfrac{35}{3} + (9 - 1)5 \\[1em] = -\dfrac{35}{3} + 40 \\[1em] = \dfrac{-35 + 120}{3} \\[1em] = \dfrac{85}{3}.$

Hence, a = $-\dfrac{35}{3} \text{ and } a_9 = \dfrac{85}{3}.$

(iv) a = 8, a_n = 62, S_n = 210.

By formula a_n = a + (n - 1)d

⇒ 62 = 8 + (n - 1)d
⇒ 62 - 8 = (n - 1)d
⇒ (n - 1)d = 54          (Eq 1)

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow 210 = \dfrac{n}{2}[2 \times 8 + (n - 1)d] \\[1em]$

Putting value of (n - 1)d from Eq 1 in above equation,

$\Rightarrow \dfrac{n}{2}[16 + 54] = 210 \\[1em] \Rightarrow \dfrac{n}{2} \times 70 = 210 \\[1em] \Rightarrow 35n = 210 \\[1em] \Rightarrow n = \dfrac{210}{35} \\[1em] \Rightarrow n = 6 \\[1em] \text{From Eq 1, } (n - 1)d = 54 \\[1em] \Rightarrow (6 - 1)d = 54 \\[1em] \Rightarrow 5d = 54 \\[1em] \therefore d = \dfrac{54}{5} \text{ and } n = 6.$

Hence, the value of n = 6 and d = $\dfrac{54}{5}.$

(v) a = 3, n = 8, S = 192.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ 192 = $\dfrac{8}{2}[2 \times 3 + (8 - 1)d]$
⇒ 192 = 4[6 + 7d]

⇒ $\dfrac{192}{4}$ = 6 + 7d
⇒ 48 = 6 + 7d
⇒ 7d = 42
⇒ d = 6.

Hence, the value of d is 6.

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

Answer

Let nth be the last term so,
a = 5, l = a_n = 45, S_n = 400.

By formula a_n = a + (n - 1)d
⇒ 45 = 5 + (n - 1)d
⇒ 45 - 5 = (n - 1)d
⇒ (n - 1)d = 40      (Eq 1)

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow 400 = \dfrac{n}{2}[2 \times 5 + (n - 1)d]$

Putting value of (n - 1)d from Eq 1 in above equation,

$\Rightarrow 400 = \dfrac{n}{2}[10 + 40] \\[1em] \Rightarrow \dfrac{n}{2} \times 50 = 400 \\[1em] \Rightarrow 25n = 400 \\[1em] \Rightarrow n = 16$

Putting value of n in Eq 1 we get,

$\Rightarrow (16 - 1)d = 40 \\[1em] \Rightarrow 15d = 40 \\[1em] \Rightarrow d = \dfrac{40}{15} = \dfrac{8}{3}. \\[1em]$

Hence, number of terms = 16 and common difference = $\dfrac{8}{3}.$

The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.

Answer

Given, a = 15, S₁₅ = 750

By formula, S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow S_{15} = \dfrac{15}{2}[2 \times 15 + (15 - 1)d] \\[1em] \Rightarrow 750 = \dfrac{15}{2}[30 + 14d] \\[1em] \Rightarrow 750 \times 2 = 450 + 210d \\[1em] \Rightarrow 450 + 210d = 1500 \\[1em] \Rightarrow 210d = 1500 - 450 \\[1em] \Rightarrow 210d = 1050 \\[1em] \Rightarrow d = \dfrac{1050}{210} \\[1em] \Rightarrow d = 5.$

By formula, a_n = a + (n - 1)d
⇒ a₂₀ = 15 + (20 - 1)5
⇒ a₂₀ = 15 + 95
⇒ a₂₀ = 110.

Hence, the 20th term of the A.P. is 110.

The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Answer

Given, a = 17, l = 350 and d = 9.

Let nth be the last term so,
l = a_n = 350

By formula, a_n = a + (n - 1)d
⇒ 350 = 17 + (n - 1)9
⇒ 350 = 17 + 9n - 9
⇒ 350 = 9n + 8
⇒ 9n = 350 - 8
⇒ 9n = 342
⇒ n = 38.

By formula, S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₃₈ = $\dfrac{38}{2}[2 \times 17 + (38 - 1)9]$
⇒ S₃₈ = 19[34 + 37(9)]
⇒ 19[34 + 333]
⇒ 19(367)
⇒ 6973.

Hence, number of terms = 38 and the sum of the terms = 6973.

Solve for x : 1 + 4 + 7 + 10 + ... + x = 287.

Answer

The above series is in A.P. because,
any term - preceding term = 3 = common difference.

Let there be n terms so, x = a_n.

Given, a = 1, d = 3 and S_n = 287.

By formula, S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow S_n = \dfrac{n}{2}[2 \times 1 + (n - 1)3] \\[1em] \Rightarrow 287 = \dfrac{n}{2}[2 + 3n - 3] \\[1em] \Rightarrow 287 \times 2 = n[3n - 1] \\[1em] \Rightarrow 3n^2 - n = 574 \\[1em] \Rightarrow 3n^2 - n - 574 = 0 \\[1em] 3n^2 - 42n + 41n - 574 = 0 \\[1em] 3n(n - 14) + 41(n - 14) = 0 \\[1em] (3n + 41)(n - 14) = 0 \\[1em] 3n + 41 = 0 \text{ or } n - 14 = 0 \\[1em] n = -\dfrac{41}{3} \text{ or } n = 14.$

Since number of terms cannot be negative so n ≠ $-\dfrac{41}{3}$

∴ n = 14.

We know x = a_n so,

⇒ x = a + (n - 1)d
⇒ x = 1 + (14 - 1)3
⇒ x = 1 + 13(3)
⇒ x = 1 + 39
⇒ x = 40.

Hence, the value of x = 40.

How many terms of the A.P. 25, 22, 19, .... are needed to give the sum 116 ? Also find the last term.

Answer

Let the number of terms required be n.

For the given A.P.,
a = 25, d = 22 - 25 = -3, S_n = 116

By formula, S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow 116 = \dfrac{n}{2}[2 \times 25 + (n - 1)(-3)] \\[1em] 116 \times 2 = n[50 - 3n + 3] \\[1em] 232 = n[53 - 3n] \\[1em] 232 = 53n - 3n^2 \\[1em] 3n^2 - 53n + 232 = 0 \\[1em] 3n^2 - 24n - 29n + 232 = 0 \\[1em] 3n(n - 8) - 29(n - 8) = 0 \\[1em] (3n - 29)(n - 8) = 0 \\[1em] 3n - 29 = 0 \text{ or } n - 8 = 0 \\[1em] n = \dfrac{29}{3} \text{ or } n = 8.$

Since, number of terms will be a natural number so n ≠ $\dfrac{29}{3}$

∴ n = 8.

By formula S_n = $\dfrac{n}{2}[a + l]$

⇒ 116 = $\dfrac{8}{2}[25 + l]$
⇒ 116 = 4[25 + l]
⇒ 29 = 25 + l
⇒ 29 - 25 = l
⇒ l = 4.

Hence, number of terms = 8 and the last term = 4.

How many terms of the A.P. 24, 21, 18, ... must be taken so that the sum is 78 ? Explain the double answer.

Answer

Let numbers of terms be n.
Given, a = 24, d = 21 - 24 = -3 and S_n = 78.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ 78 = $\dfrac{n}{2}[2 \times 24 + (n - 1)(-3)]$
⇒ 78 × 2 = n[48 - 3n + 3]
⇒ 156 = n[51 - 3n]
⇒ 156 = 51n - 3n²
⇒ 3n² - 51n + 156 = 0
⇒ 3n² - 12n - 39n + 156 = 0
⇒ 3n(n - 4) - 39(n - 4) = 0
⇒ (3n - 39)(n - 4) = 0
⇒ 3n - 39 = 0 or n - 4 = 0
⇒ 3n = 39 or n = 4
⇒ n = 13 or n = 4.

Let's check sum of 4 terms
⇒ a₄ = a₃ + d
⇒ a₄ = 18 + (-3) = 15

∴ S₄ = 24 + 21 + 18 + 15 = 78.

By formula, a_n = a + (n - 1)d
⇒ a₅ = 24 + (5 - 1)(-3)
⇒ a₅ = 24 + 4(-3)
⇒ a₅ = 24 - 12 = 12.

a₆ = a₅ + d = 12 + (-3) = 9
a₇ = a₆ + d = 9 + (-3) = 6
a₈ = a₇ + d = 6 + (-3) = 3
a₉ = a₈ + d = 3 + (-3) = 0
a₁₀ = a₉ + d = 0 + (-3) = -3
a₁₁ = a₁₀ + d = -3 + (-3) = -6
a₁₂ = a₁₁ + d = -6 + (-3) = -9
a₁₃ = a₁₂ + d = -9 + (-3) = -12.

Taking sum of from 5th term to 13th,

12 + 9 + 6 + 3 + 0 - 3 - 6 - 9 - 12 = 0

Since, the sum from 5tn term to 13th is zero hence, it will not add a value.

∴ n = 13 and 4.

Hence, the number of terms that can be taken for sum to be 78 can be 4 or 13.

Find the sum of first 22 terms of an A.P. in which d = 7 and a₂₂ is 149.

Answer

Given, d = 7 and a₂₂ = 149.

By formula, a_n = a + (n - 1)d

⇒ 149 = a + (22 - 1)7
⇒ 149 = a + 21(7)
⇒ 149 = a + 147
⇒ a = 149 - 147
⇒ a = 2.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S_n = $\dfrac{22}{2}[2 \times 2 + (22 - 1)7]$
⇒ S_n = 11[4 + 147]
⇒ S_n = 11 × 151
⇒ S_n = 1661.

Hence, the sum of first 22 terms of the A.P. is 1661.

In an A.P., the 5^th and 9^th term are 4 and -12 respectively. Find :

(a) the first term

(b) common difference

(c) sum of the first 20 terms.

Answer

If first term is a and common difference is d of the A.P.

By formula,

n^th term = a_n = a + (n - 1)d

Given,

⇒ 5^th term = 4

⇒ a₅ = 4

⇒ a + (5 - 1)d = 4

⇒ a + 4d = 4 .........(1)

⇒ 9^th term = -12

⇒ a₉ = -12

⇒ a + (9 - 1)d = -12

⇒ a + 8d = -12 .........(2)

Subtracting equation (1) from (2), we get :

⇒ (a + 8d) - (a + 4d) = -12 - 4

⇒ a - a + 8d - 4d = -16

⇒ 4d = -16

⇒ d = $-\dfrac{16}{4}$

⇒ d = -4.

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

⇒ a + 4d = 4

⇒ a + 4(-4) = 4

⇒ a - 16 = 4

⇒ a = 4 + 16 = 20.

(a) Hence, first term of A.P. = 20.

(b) Common difference of A.P. = -4.

(c) By formula,

Sum of n terms of A.P. = $\dfrac{n}{2}(a + l)$

Sum of first 20 terms of A.P. = $\dfrac{n}{2}(a + a_{20})$

$= \dfrac{20}{2}[a + a + (20 - 1)d] \\[1em] = 10(2a + 19d) \\[1em] = 10 \times (2 \times 20 + 19 \times -4) \\[1em] = 10 \times (40 - 76) \\[1em] = 10 \times -36 \\[1em] = -360.$

Hence, sum of first 20 terms of A.P. = -360.

Find the sum of first 51 terms of the A.P. whose second and third terms are 14 and 18 respectively.

Answer

Given, a₂ = 14 and a₃ = 18.

common difference = d = any term - preceding term = a₃ - a₂ = 18 - 14 = 4.

By formula, a_n = a + (n - 1)d
⇒ a₂ = a + 4(2 - 1)
⇒ 14 = a + 4
⇒ a = 10.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow S_{51} = \dfrac{51}{2}[2 \times 10 + (51 - 1)4] \\[1em] = \dfrac{51}{2}[20 + 50 \times 4] \\[1em] = \dfrac{51}{2}[20 + 200] \\[1em] = \dfrac{51}{2} \times 220 \\[1em] = 51 \times 110 \\[1em] = 5610.$

Hence, the sum of first 51 terms of the A.P. is 5610.

The 4th term of an A.P. is 22 and 15th term is 66. Find the first term and the common difference. Hence, find the sum of first 8 terms of the A.P.

Answer

Given, a₄ = 22 and a₁₅ = 66.

By formula, a_n = a + (n - 1)d
⇒ a₄ = a + (4 - 1)d
⇒ 22 = a + 3d
⇒ a = 22 - 3d         (Eq 1)

⇒ a₁₅ = a + (15 - 1)d
⇒ 66 = a + 14d
Putting value of a from Eq 1 in above equation
⇒ 66 = 22 - 3d + 14d
⇒ 66 = 22 + 11d
⇒ 11d = 66 - 22
⇒ 11d = 44
⇒ d = 4.

Putting value of d in Eq 1,
⇒ a = 22 - 3(4)
⇒ a = 22 - 12
⇒ a = 10.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₈ = $\dfrac{8}{2}[2 \times 10 + (8 - 1)4]$
⇒ S₈ = 4[20 + 28]
⇒ S₈ = 4 × 48
⇒ S₈ = 192.

Hence, first term = a = 10, common difference = d = 4 and sum of first 8 terms = S₈ = 192.

If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

Answer

Given, S₆ = 36 and S₁₆ = 256.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₆ = $\dfrac{6}{2}[2a + (6 - 1)d]$
⇒ 36 = 3[2a + 5d]
⇒ 2a + 5d = 12
⇒ 2a = 12 - 5d     (Eq 1)

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₁₆ = $\dfrac{16}{2}[2a + (16 - 1)d]$
⇒ 256 = 8[2a + 15d]
⇒ 2a + 15d = 32
⇒ 2a = 32 - 15d

Putting value of 2a from Eq 1 in above equation,

⇒ 12 - 5d = 32 - 15d
⇒ -5d + 15d = 32 - 12
⇒ 10d = 20
⇒ d = 2.

∴ From Eq 1,
⇒ 2a = 12 - 5d
⇒ 2a = 12 - 5(2)
⇒ 2a = 2
⇒ a = 1.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₁₀ = $\dfrac{10}{2}[2 \times 1 + (10 - 1)2]$
⇒ S₁₀ = 5[2 + 18]
⇒ S₁₀ = 5 × 20
⇒ S₁₀ = 100.

Hence, the sum of first 10 terms of the A.P. is 100.

Show that a₁, a₂, a₃, ..... form an A.P. where a_n is defined as a_n = 3 + 4n. Also find the sum of first 15 terms.

Answer

a_n = 3 + 4n
a₁ = 3 + 4 × 1 = 3 + 4 = 7
a₂ = 3 + 4 × 2 = 3 + 8 = 11
a₃ = 3 + 4 × 3 = 3 + 12 = 15
a₄ = 3 + 4 × 4 = 3 + 16 = 19.

Since, a₄ - a₃ = a₃ - a₂ = a₂ - a₁ = 4, i.e any term - preceding term = fixed number = 4.

∴ a₁, a₂, a₃ ..... form an A.P.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow S_{15} = \dfrac{15}{2}[2 \times 7 + (15 - 1)4] \\[1em] = \dfrac{15}{2}[14 + 14 \times 4] \\[1em] = \dfrac{15}{2}[14 + 56] \\[1em] = \dfrac{15}{2} \times 70 \\[1em] = 15 \times 35 \\[1em] = 525.$

Hence, the sum of first 15 terms of the A.P. is 525.

The sum of first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is 1 : 3. Calculate the first and the thirteenth term.

Answer

Given, S₆ = 42 and a₁₀ : a₃₀ = 1 : 3

$\Rightarrow a_{10} : a_{30} = 1 : 3 \\[1em] \Rightarrow \dfrac{a_{10}}{a_{30}} = \dfrac{1}{3} \\[1em] \Rightarrow 3a_{10} = a_{30}$

By formula, a_n = a + (n - 1)d,

$\Rightarrow 3[a + (10 - 1)d] = a + (30 - 1)d \\[1em] \Rightarrow 3(a + 9d) = a + 29d \\[1em] \Rightarrow 3a + 27d = a + 29d \\[1em] \Rightarrow 3a - a = 29d - 27d \\[1em] \Rightarrow 2a = 2d \\[1em] \Rightarrow a = d.$

S₆ = 42 or,

⇒ $\dfrac{6}{2}[2a + (6 - 1)d]$ = 42

Since, a = d
⇒ 3[2d + 5d] = 42
⇒ 3 × 7d = 42
⇒ 21d = 42
⇒ d = 2.

Since, a = d hence a = 2.

By formula, a_n = a + (n - 1)d
⇒ a₁₃ = 2 + (13 - 1)(2)
⇒ a₁₃ = 2 + 24
⇒ a₁₃ = 26.

Hence, the first term of the A.P. is 2 and the thirteenth term is 26.

In an A.P., the sum of its first n terms is 6n - n². Find its 25th term.

Answer

Given, S_n = 6n - n².

So, sum of (n -1) terms will be

⇒ S_{n - 1} = 6(n - 1) - (n - 1)²
⇒ S_{n - 1} = 6n - 6 -(n² + 1 - 2n)
⇒ S_{n - 1} = 6n - 6 - n² - 1 + 2n
⇒ S_{n - 1} = 8n - n² - 7.

By formula, a_n = S_n - S_{n - 1}
⇒ a_n = 6n - n² - (8n - n² - 7)
⇒ a_n = 6n - 8n - n² + n² + 7
⇒ a_n = -2n + 7.

∴ a₂₅ = -2(25) + 7 = -50 + 7 = -43.

Hence, the 25th term of the A.P. is -43.

If S_n denotes the sum of first n terms of an A.P., prove that S₃₀ = 3(S₂₀ - S₁₀).

Answer

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

We need to prove S₃₀ = 3(S₂₀ - S₁₀).

$\Rightarrow \text{L.H.S.} = S_{30} = \dfrac{30}{2}[2a + (30 - 1)d] \\[1em] = 15[2a + 29d] \\[1em] = 30a + 435d \\[1em] \Rightarrow \text{R.H.S.} = 3(S_{20} - S_{10}) \\[1em] = 3\Big(\dfrac{20}{2}[2a + (20 - 1)d] - \dfrac{10}{2}[2a + (10 - 1)d]\Big) \\[1em] = 3\Big(10[2a + 19d] - 5[2a + 9d]\Big) \\[1em] = 3(20a + 190d - 10a - 45d) \\[1em] = 3(10a + 145d) \\[1em] = 30a + 435d.$

∴ L.H.S. = R.H.S. = 30a + 435d.

Hence, proved that S₃₀ = 3(S₂₀ - S₁₀).

Find the sum of first positive 1000 integers.

Answer

We need to find 1 + 2 + 3 + 4 + ...... + 1000

The above series is an A.P. with first term a = 1, d = 1 and n = 1000.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₁₀₀₀ = $\dfrac{1000}{2}[2 \times 1 + (1000 - 1)1]$
⇒ S₁₀₀₀ = 500[2 + 999]
⇒ S₁₀₀₀ = 500 × 1001
⇒ S₁₀₀₀ = 500500.

Hence, the sum of first 1000 positive integers is 500500.

Find the sum of first 15 multiples of 8.

Answer

We need to find 8 + 16 + 24 + ..... + (15th term)

The above series forms an A.P. with a = 8 and d = 8.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₁₅ = $\dfrac{15}{2}[2 \times 8 + (15 - 1)8]$
⇒ 2S₁₅ = 15[16 + 14(8)]
⇒ 2S₁₅ = 15[128]
⇒ 2S₁₅ = 1920
⇒ S₁₅ = 960

Hence, the sum of first 15 terms is 960.

Find the sum of all two digit natural numbers which are divisible by 4.

Answer

The sum of series of the two digit natural numbers that are divisible by 4 is $12 + 16 + 20 + .... + 96.$

a = 12, d = 16 - 12 = 4 and l = 96.

Let 96 be nth term then,

⇒ 96 = 12 + 4(n - 1)
⇒ 96 - 12 = 4n - 4
⇒ 84 = 4n - 4
⇒ 4n = 84 + 4
⇒ 4n = 88
⇒ n = 22.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₂₂ = $\dfrac{22}{2}[2 \times 12 + 4(22 - 1)]$
⇒ S₂₂ = 11[24 + 84]
⇒ S₂₂ = 11 × 108
⇒ S₂₂ = 1188.

Hence, the sum of all two digits natural numbers which are divisible by 4 is 1188.

Find the sum of all natural numbers between 100 and 200 which are divisible by 4.

Answer

The sum of natural numbers between 100 and 200 that are divisible by 4 is 104 + 108 + 112 + ..... + 196.

The above series is an A.P. with a = 104, d = 108 - 104 = 4 and l = 196.

Let 196 be nth term then,

⇒ 196 = 104 + 4(n - 1)
⇒ 196 - 104 = 4n - 4
⇒ 92 = 4n - 4
⇒ 4n = 92 + 4
⇒ 4n = 96
⇒ n = 24.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₂₄ = $\dfrac{24}{2}[2 \times 104 + 4(24 - 1)]$
⇒ S₂₄ = 12[208 + 92]
⇒ S₂₄ = 12 × 300
⇒ S₂₄ = 3600.

Hence, the sum of all two digits natural numbers between 100 and 200 which are divisible by 4 is 3600.

Find the sum of all multiples of 9 lying between 300 and 700.

Answer

The sum of all multiples of 9 lying between 300 and 700 is 306 + 315 + ..... + 693.

The above series is an A.P. with a = 306, d = 9 and l = 693.

Let 693 be nth term of the series then,

⇒ 693 = 306 + 9(n - 1)
⇒ 693 - 306 = 9n - 9
⇒ 387 = 9n - 9
⇒ 9n = 387 + 9
⇒ 9n = 396
⇒ n = 44.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₄₄ = $\dfrac{44}{2}[2 \times 306 + 9(44 - 1)]$
⇒ S₄₄ = 22[612 + 387]
⇒ S₄₄ = 22 × 999
⇒ S₄₄ = 21978.

Hence, the sum of all multiples of 9 lying between 300 and 700 is 21978.

Find the sum of all natural numbers less than 100 which are divisible by 6.

Answer

The sum of all natural numbers less than 100 which are divisible by 6 is given as 6 + 12 + 18 + .... + 96.

The above series is an A.P. with a = 6, d = 6 and l = 96.

Let 96 be nth term of the series then,

⇒ 96 = 6 + 6(n - 1)
⇒ 96 - 6 = 6n - 6
⇒ 90 = 6n - 6
⇒ 6n = 90 + 6
⇒ 6n = 96
⇒ n = 16.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₁₆ = $\dfrac{16}{2}[2 \times 6 + 6(16 - 1)]$
⇒ S₁₆ = 8[12 + 90]
⇒ S₁₆ = 8 × 102
⇒ S₁₆ = 816.

Hence, the sum of all natural numbers less than 100 which are divisible by 6 is 816.

An arithmetic progression (A.P.) has 3 as its first term. The sum of the first 8 terms is twice the sum of the first 5 terms. Find the common difference of the A.P.

Answer

Let common difference be d.

a = 3

Sum of first n terms of an A.P. = $\dfrac{\text{n}}{2}(\text{a } + \text{ l})$

Given,

The sum of the first 8 terms is twice the sum of the first 5 terms.

$\therefore \dfrac{8}{2}(a + a_8) = 2 \times \dfrac{5}{2}(a + a_5) \\[1em] \Rightarrow 4[a + a + (8 - 1)d] = 5[a + a + (5 - 1)d] \\[1em] \Rightarrow 4[2a + 7d] = 5[2a + 4d] \\[1em] \Rightarrow 4[2 \times 3 + 7d] = 5[2 \times 3 + 4d] \\[1em] \Rightarrow 4[6 + 7d] = 5[6 + 4d] \\[1em] \Rightarrow 24 + 28d = 30 + 20d \\[1em] \Rightarrow 28d - 20d = 30 - 24 \\[1em] \Rightarrow 8d = 6 \\[1em] \Rightarrow d = \dfrac{6}{8} = \dfrac{3}{4}.$

Hence, common difference = $\dfrac{3}{4}$.