Arithmetic Progression — Exercise 10(D)

k + 2, 2k + 7 and 4k + 12 are the first three terms of an A.P. The first term of this A.P. is :

1. -2

2. 0

3. 2

4. 3

Answer

Since,

k + 2, 2k + 7 and 4k + 12 are the first three terms of an A.P.

∴ 2k + 7 - (k + 2) = 4k + 12 - (2k + 7)

⇒ 2k - k + 7 - 2 = 4k - 2k + 12 - 7

⇒ k + 5 = 2k + 5

⇒ 2k - k = 5 - 5

⇒ k = 0.

⇒ k + 2 = 0 + 2 = 2.

Hence, Option 3 is the correct option.

The sum of n terms of an A.P. is 3n². The second term of this A.P. is :

1. 8

2. 3

3. 9

4. 12

Answer

Given,

Sum of n terms = 3n²

Sum upto 2 terms = 3(2)² = 3 × 4 = 12.

Sum upto 1 term = 3(1)² = 3 × 1 = 3.

2nd term = Sum upto 2 terms - Sum upto 1 terms = 12 - 3 = 9.

Hence, Option 3 is the correct option.

If 5, 7 and 9 are in A.P. then which of the following is in A.P.?

1. 5 × 7, 7 × 9 and 9 × 5

2. 5 × 7, 7 × 7 and 9 × 7

3. 2 × 5, 2 × 7 and 5 × 9

4. 5 - 7, 7 - 9 and 9 - 5

Answer

First list :

⇒ 5 × 7, 7 × 9 and 9 × 5

⇒ 35, 63, 45

63 - 35 = 28

45 - 63 = -18

Since, common difference between consecutive terms is not equal, so it is not an A.P.

Second list :

⇒ 5 × 7, 7 × 7 and 9 × 7

⇒ 35, 49, 63

49 - 35 = 14

63 - 49 = 14

Since, common difference between consecutive terms are equal, it is an A.P.

Hence, Option 2 is the correct option.

Find three numbers in A.P. whose sum is 24 and whose product is 440.

Answer

Let three numbers in A.P. be (a - d), a, (a + d).

Sum = 24

∴ a - d + a + a + d = 24

⇒ 3a = 24

⇒ a = 8.

Product = 440

⇒ (a - d)(a)(a + d) = 440

⇒ (8 - d)(8)(8 + d) = 440

⇒ (8 - d)(8 + d) = $\dfrac{440}{8}$

⇒ 64 - d² = 55

⇒ d² = 64 - 55 = 9

⇒ d = ± 3

Let d = 3,

A.P. = (8 - 3), 8, (8 + 3) = 5, 8, 11.

Let d = -3,

A.P. = (8 - (-3)), 8, (8 + (-3)) = 11, 8, 5.

Hence, A.P. = 5, 8, 11 or 11, 8, 5.

The angles of a quadrilateral are in A.P. with common difference 20°. Find its angles.

Answer

Let the angles of quadrilateral are,

a, a + d, a + 2d, a + 3d

∴ a + (a + d) + (a + 2d) + (a + 3d) = 360°

⇒ 4a + 6d = 360°

⇒ 2(2a + 3d) = 360°

⇒ 2a + 3d = 180°

Putting value of d = 20° in above equation we get,

⇒ 2a + 3(20) = 180°

⇒ 2a + 60 = 180°

⇒ 2a = 180° - 60 = 120°

⇒ a = 60°.

Hence, angles = 60°, 80°, 100°, 120°.

Divide 96 into four parts which are in A.P. and the ratio between product of their means to product of their extremes is 15 : 7.

Answer

Let the four terms of A.P. are

(a - 3d), (a - d), (a + d), (a + 3d).

According to question,

⇒ a - 3d + a - d + a + d + a + 3d = 96

⇒ 4a = 96

⇒ a = 24.

So, terms are,

24 - 3d, 24 - d, 24 + d, 24 + 3d.

Given, ratio between product of their means to product of their extremes is 15 : 7.

$\therefore \dfrac{(24 - d)(24 + d)}{(24 - 3d)(24 + 3d)} = \dfrac{15}{7} \\[1em] \Rightarrow \dfrac{576 - d^2}{576 - 9d^2} = \dfrac{15}{7} \\[1em] \Rightarrow 7(576 - d^2) = 15(576 - 9d^2) \\[1em] \Rightarrow 4032 - 7d^2 = 8640 - 135d^2 \\[1em] \Rightarrow 135d^2 - 7d^2 = 8640 - 4032 \\[1em] \Rightarrow 128d^2 = 4608 \\[1em] \Rightarrow d^2 = 36 \\[1em] \Rightarrow d = \pm 6.$

Let d = 6,

Terms = (a - 3d), (a - d), (a + d), (a + 3d) = (24 - 3 × 6), (24 - 6), (24 + 6), (24 + 3 × 6)

= 6, 18, 30, 42.

Let d = -6,

Terms = (a - 3d), (a - d), (a + d), (a + 3d) = (24 - 3 × -6), (24 - (-6)), (24 + (-6)), (24 + 3 × -6)

= 42, 30, 18, 6.

Hence, four parts of 96 are 6, 18, 30, 42 or 42, 30, 18, 6.

Find five numbers in A.P. whose sum is $12\dfrac{1}{2}$ and the ratio of the first to the last term is 2 : 3.

Answer

Let the numbers be (a - 2d), (a - d), a, (a + d), (a + 2d).

Given, sum = $12\dfrac{1}{2}$.

$\therefore a - 2d + a - d + a + a + d + a + 2d = 12\dfrac{1}{2} \\[1em] \Rightarrow 5a = \dfrac{25}{2} \\[1em] \Rightarrow a = \dfrac{5}{2} = 2.5.$

Given, ratio of the first to the last term is 2 : 3.

$\therefore \dfrac{a - 2d}{a + 2d} = \dfrac{2}{3} \\[1em] \Rightarrow 3(a - 2d) = 2(a + 2d) \\[1em] \Rightarrow 3a - 6d = 2a + 4d \\[1em] \Rightarrow 3a - 2a = 4d + 6d \\[1em] \Rightarrow a = 10d \\[1em] \Rightarrow \dfrac{5}{2} = 10d \\[1em] \Rightarrow d = \dfrac{5}{2 \times 10} \\[1em] \Rightarrow d = \dfrac{1}{4} = 0.25.$

Numbers = 2.5 - 2 × 0.25, 2.5 - 0.25, 2.5, 2.5 + 0.25, 2.5 + 2 × 0.25

= 2, 2.25, 2.5, 2.75, 3.

Hence, the numbers are 2, 2.25, 2.5, 2.75, 3.

Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts is 4623.

Answer

Let the numbers be a - d, a, a + d.

According to question,

⇒ a - d + a + a + d = 207

⇒ 3a = 207

⇒ a = 69.

Given,

Product of the two smaller parts is 4623

⇒ (a - d)(a) = 4623

⇒ 69(69 - d) = 4623

⇒ 4761 - 69d = 4623

⇒ 69d = 4761 - 4623

⇒ 69d = 138

⇒ d = 2.

Numbers = (69 - 2), 69, (69 + 2).

Hence, numbers are 67, 69, 71 or 71, 69, 67.

The sum of three numbers in A.P. is 15 and the sum of the squares of the extreme terms is 58. Find the numbers.

Answer

Let the numbers be a - d, a, a + d.

According to question,

⇒ a - d + a + a + d = 15

⇒ 3a = 15

⇒ a = 5.

Given,

Sum of the squares of the extreme terms is 58.

⇒ (a - d)² + (a + d)² = 58

⇒ (5 - d)² + (5 + d)² = 58

⇒ 25 + d² - 10d + 25 + d² + 10d = 58

⇒ 50 + 2d² = 58

⇒ 2d² = 8

⇒ d² = 4

⇒ d = ±2

Let d = 2,

Numbers = (5 - 2), 5, (5 + 2) = 3, 5, 7.

Let d = -2,

Numbers = (5 - (-2)), 5, (5 + (-2)) = 7, 5, 3.

Hence, numbers = 3, 5, 7 or 7, 5, 3.

Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.

Answer

Let numbers be a - 3d, a - d, a + d, a + 3d.

Given, sum = 20

⇒ a - 3d + a - d + a + d + a + 3d = 20

⇒ 4a = 20

⇒ a = 5.

Given, sum of squares is 120.

⇒ (a - 3d)² + (a - d)² + (a + d)² + (a + 3d)² = 120

⇒ (5 - 3d)² + (5 - d)² + (5 + d)² + (5 + 3d)² = 120

⇒ 25 + 9d² - 30d + 25 + d² - 10d + 25 + d² + 10d + 25 + 9d² + 30d = 120

⇒ 100 + 20d² = 120

⇒ 20d² = 20

⇒ d² = 1.

⇒ d = ±1

Let d = 1,

Numbers = (5 - 3(1)), (5 - 1), (5 + 1), (5 + 3(1))

= 2, 4, 6, 8.

Let d = -1,

Numbers = (5 - 3(-1)), (5 - (-1)), (5 + (-1)), (5 + 3(-1))

= 8, 6, 4, 2.

Hence, numbers are 2, 4, 6, 8 or 8, 6, 4, 2.

Insert one arithmetic mean between 3 and 13.

Answer

Arithmetic mean between 3 and 13 = $\dfrac{3 + 13}{2} = \dfrac{16}{2}$ = 8.

Hence, arithmetic mean between 3 and 13 = 8.

The angles of a polygon are in A.P. with common difference 5°. If the smallest angle is 120°, find the number of sides of the polygon.

Answer

Given, angles of polygon are in A.P.,

a = 120° and d = 5°.

Let no. of sides be n and so sum of angles = (2n - 4) × 90°

Sum of A.P. of angles = $\dfrac{n}{2}[2 \times 120° + (n - 1) \times 5°]$

$\therefore \dfrac{n}{2}[2 \times 120° + (n - 1) \times 5°] = (2n - 4) × 90° \\[1em] \Rightarrow \dfrac{n}{2}[240° + 5°n - 5°] = 180°n - 360° \\[1em] \Rightarrow 240°n + 5°n^2 - 5°n = 360°n - 720° \\[1em] \Rightarrow 5°n^2 + 235°n = 360°n - 720° \\[1em] \Rightarrow 5°n^2 - 125°n + 720° = 0 \\[1em] \Rightarrow 5°(n^2 - 25n + 144) = 0 \\[1em] \Rightarrow n^2 - 25n + 144 = 0 \\[1em] \Rightarrow n^2 - 16n - 9n + 144 = 0 \\[1em] \Rightarrow n(n - 16) - 9(n - 16) = 0 \\[1em] \Rightarrow (n - 9)(n - 16) = 0 \\[1em] \Rightarrow n = 9, 16.$

Hence, number of sides = 9 or 16.

If p^th term of an A.P. is q and its q^th term is p, show that its (p + q)^th term is zero.

Answer

Let the first term be a and common difference be d.

Then,

T_p = a + (p − 1)d = q …(1)

T_q = a + (q − 1)d = p …(2)

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

⇒ a + (p − 1)d − [a + (q − 1)d] = q − p

⇒ (p − 1)d − (q − 1)d = q − p

⇒ pd - d - qd + d = q - p

⇒ pd - qd = q - p

⇒ (p − q)d = q − p

⇒ (p − q)d = −(p − q)

⇒ d = $\dfrac{−(p − q)}{(p − q)}$

⇒ d = -1.

Substituting d = −1 in equation (1), we get :

⇒ a + (p − 1)(−1) = q

⇒ a − p + 1 = q

⇒ a = p + q − 1.

Now, the (p + q)^th term :

T_{p + q} = a + (p + q − 1)d

= (p + q − 1) + (p + q − 1)(−1)

= (p + q − 1) − (p + q − 1)

= p - p + q - q - 1 + 1

= 0.

Hence, the (p + q)^th term of the A.P. is zero.

If a, b and c are p^th, q^th and r^th terms of an A.P., prove that

a(q - r) + b(r − p) + c(p − q) = 0

Answer

Let t and d be the first term and common difference of the A.P respectively.

The nth term of an A.P is given by, a_n = t + (n - 1) d

pth term = t + (p - 1)d = a ....(1)

qth term = t + (q - 1)d = b ....(2)

rth term = t + (r - 1)d = c ....(3)

Subtracting (2) from (1), we obtain :

⇒ t + (p - 1)d - [t + (q - 1)d] = a - b

⇒ t + (p - 1)d - t - (q - 1)d = a - b

⇒ (p - 1 - q + 1)d = a - b

⇒ (p - q)d = a - b

⇒ d = $\dfrac{(a - b)}{(p - q)}$ ....(4)

Subtracting (3) from (2), we obtain :

⇒ t + (q - 1)d - [t + (r - 1)d] = b - c

⇒ t + (q - 1)d - t - (r - 1) d = b - c

⇒ (q - 1 - r + 1)d = b - c

⇒ (q - r)d = b - c

⇒ d = $\dfrac{b - c}{q - r}$ ....(5)

From (4) and (5), we get:

$\dfrac{(a - b)}{(p - q)} = \dfrac{b - c}{q - r}$

⇒ (a - b)(q - r ) = (b - c)( p - q)

⇒ aq - ar - bq + br = bp - bq - cp + cq

⇒ bp - cp + cq - aq + ar - br - bq + bq = 0

⇒ (-aq + ar) + (bp - br) + (-cp + cq) = 0

⇒ -a(q - r) - b(r - p) - c(p - q) = 0

⇒ a(q - r) + b(r - p) + c(p - q) = 0

Hence, proved that a(q - r) + b(r − p) + c(p − q) = 0.

Show that a², b² and c² are in A.P., if

$\dfrac{1}{b + c}, \dfrac{1}{c + a} \text{ and } \dfrac{1}{a + b}$ are in A.P.

Answer

Since $\dfrac{1}{b + c}, \dfrac{1}{c + a}, \dfrac{1}{a + b}$ are in A.P.,

$\Rightarrow \dfrac{1}{c + a} - \dfrac{1}{b + c} = \dfrac{1}{a + b} - \dfrac{1}{c + a} \\[1em] \Rightarrow \dfrac{(b + c) - (c + a)}{(c + a)(b + c)} = \dfrac{(c + a) - (a + b)}{(a + b)(c + a)} \\[1em] \Rightarrow \dfrac{b - a}{(c + a)(b + c)} = \dfrac{c - b}{(a + b)(c + a)} \\[1em] \Rightarrow \dfrac{b - a}{b + c} = \dfrac{c - b}{a + b} \\[1em]$

⇒ (b − a)(a + b) = (c − b)(b + c)

⇒ ab + b² − a² − ab = bc + c² − b² − bc

⇒ b² − a² = c² − b²

⇒ b² − a² = c² − b²

We have,

b² − a² = c² − b²

Since consecutive differences are equal,

Therefore,

a², b², c² are in A.P.

Hence, a², b², c² are in A.P.

A man saved ₹ 7,65,000 in 10 years. In each year, after the first, he saved ₹ 6,000 more than he did in the preceding year. How much did he save in the first seven years?

Answer

Total saving in 10 years = ₹ 7,65,000

Each year he saved ₹ 6,000 more than previous year

d = 6000

Let his first year saving be ₹ a.

Number of years (n) = 10

By formula,

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

Substituting values, we get :

$\Rightarrow 765000 = \dfrac{10}{2}[2a + 9(6000)] \\[1em] \Rightarrow 765000 = 5[2a + 54000] \\[1em] \Rightarrow 765000 = 10a + 270000 \\[1em] \Rightarrow 10a = 765000 - 270000 \\[1em] \Rightarrow 10a = 495000 \\[1em] \Rightarrow a = \dfrac{495000}{10} \\[1em] \Rightarrow a = 49500$

Savings in the first seven years :

$\Rightarrow S_7 = \dfrac{7}{2}[2(49500) + (7 - 1)(6000)] \\[1em] = 3.5[99000 + 6(6000)] \\[1em] = 3.5(99000 + 36000) \\[1em] = 3.5(135000) \\[1em] = ₹ 4,72,500.$

Hence, amount saved in first 7 years = ₹ 4,72,500

If S_n denotes the sum of first n terms of an A.P., prove that :

S₁₂ = 3(S₈ − S₄)

Answer

By formula,

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

where, a = first term, d = common difference

$\Rightarrow S_{12} = \dfrac{12}{2}[2a + 11d] \\[1em] \Rightarrow S_{12} = 6(2a + 11d) \\[1em] \Rightarrow S_{12} = 12a + 66d \text{ ....(1)} \\[1em] \Rightarrow S_{8} = \dfrac{8}{2}[2a + 7d] \\[1em] \Rightarrow S_{8} = 4(2a + 7d) \\[1em] \Rightarrow S_{8} = 8a + 28d \text{ ....(2)} \\[1em] \Rightarrow S_{4} = \dfrac{4}{2}[2a + 3d] \\[1em] \Rightarrow S_{4} = 2(2a + 3d) \\[1em] \Rightarrow S_{4} = 4a + 6d \text{ ....(3)}$

Subtracting equation (3) from equation (2), we get :

⇒ S₈ - S₄ = (8a + 28d) - (4a + 6d)

⇒ S₈ - S₄ = 8a + 28d - 4a - 6d

⇒ S₈ - S₄ = 4a + 22d

Multiplying the above equation by 3, we get :

⇒ 3(S₈ - S₄) = 3(4a + 22d)

⇒ 3(S₈ - S₄) = 12a + 66d

⇒ 3(S₈ - S₄) = S₁₂

Hence, proved that S₁₂ = 3(S₈ − S₄).

18^th term of an A.P. is equal to 4 times its 4^th term and the 6^th term exceeds twice the 2^nd term by 4. Find the sum of the first 9 terms of this A.P.

Answer

We know that,

a_n = a + (n - 1)d

Given,

18^th term of an A.P. is equal to 4 times its 4^th term

a₁₈ = a + 17d

a₄ = a + 3d

⇒ a + 17d = 4(a + 3d)

⇒ a + 17d = 4a + 12d

⇒ a - 4a + 17d - 12d = 0

⇒ -3a + 5d = 0

⇒ 5d = 3a

⇒ a = $\dfrac{5d}{3}$ .....(1)

Given,

6th term exceeds twice 2nd term by 4

a₆ = a + 5d

a₂ = a + d

⇒ a + 5d = 2(a + d) + 4

⇒ a + 5d = 2a + 2d + 4

⇒ a + 5d - 2a - 2d = 4

⇒ 3d - a = 4

⇒ 3d - a = 4 ...(2)

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

⇒ 3d - $\dfrac{5d}{3}$ = 4

⇒ $\dfrac{9d - 5d}{3}$ = 4

⇒ 4d = 12

⇒ d = $\dfrac{12}{4}$

⇒ d = 3

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

⇒ a = $\dfrac{5d}{3}$

⇒ a = $\dfrac{5(3)}{3}$

⇒ a = 5

By formula,

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

where, a = first term, d = common difference

$\Rightarrow S_{9} = \dfrac{9}{2}[2(5) + 8(3)] \\[1em] = \dfrac{9}{2}[10 + 24] \\[1em] = \dfrac{9}{2}[34] \\[1em] = 9 \times 17 \\[1em] = 153.$

Hence, the sum of the first 9 terms of this A.P. = 153.