Arithmetic & Geometric Progression — Exercise 9.4

Find the next term of the list of numbers $\dfrac{1}{6}, \dfrac{1}{3}, \dfrac{2}{3}, ...$

Answer

The given list of numbers $\dfrac{1}{6}, \dfrac{1}{3}, \dfrac{2}{3}, ...$ is a G.P. with first term a = $\dfrac{1}{6}$ and common ratio = r = 2.

By formula, a_n = ar^{n - 1}

∴ a₄ = $\dfrac{1}{6}(2)^3 = \dfrac{8}{6} = \dfrac{4}{3}.$

Hence, the next term of the G.P. is $\dfrac{4}{3}$.

Find the next term of the list of numbers $\dfrac{3}{16}, -\dfrac{3}{8}, \dfrac{3}{4}, -\dfrac{3}{2}, ...$

Answer

The given list of numbers,

$\dfrac{3}{16}, -\dfrac{3}{8}, \dfrac{3}{4}, -\dfrac{3}{2}, ...$ is a G.P with first term a = $\dfrac{3}{16}$ and common ratio = r = -2.

By formula, a_n = ar^{n - 1}

∴ a₅ = $\dfrac{3}{16}(-2)^4 = \dfrac{3}{16} \times 16 = 3.$

Hence, the next term of the G.P. is 3.

Find the 15th term of the series $\sqrt{3} + \dfrac{1}{\sqrt{3}} + \dfrac{1}{3\sqrt{3}} + ....$

Answer

The given list of numbers,

$\sqrt{3} + \dfrac{1}{\sqrt{3}} + \dfrac{1}{3\sqrt{3}} + ....$ is a G.P. with first term a = $\sqrt{3}$ and common ratio = r = $\dfrac{1}{3}.$

By formula, a_n = ar^{n - 1}

∴ a₁₅ = $\sqrt{3}\Big(\dfrac{1}{3}\Big)^{14} = 3^{1/2}\Big(\dfrac{1}{3^{14}}\Big) = 3^{1/2}\times 3^{-14} = 3^{1/2 - 14} = 3^{-27/2} .$

Hence, the 15th term of the G.P. is $3^{-27/2}$.

Find the 10th and nth terms of the list of numbers 5, 25, 125, ...

Answer

The given list of numbers 5, 25, 125 .... is a G.P. with first term a = 5 and r = 5.

By formula, a_n = ar^{n - 1}

$a_{10} = 5(5)^{10 - 1} = 5(5)^9 = 5^{10}. \\[1em] a_n = 5(5)^{n - 1} = 5 \times 5^n \times 5^{-1} = 5^n.$

Hence, the 10th term is 5¹⁰ and nth term is 5ⁿ.

Find the 6th and the nth terms of the list of numbers $\dfrac{3}{2}, \dfrac{3}{4}, \dfrac{3}{8}, ...$

Answer

The given list of numbers

$\dfrac{3}{2}, \dfrac{3}{4}, \dfrac{3}{8}, ...$ is a G.P. with first term a = $\dfrac{3}{2}$ and common ratio = r = $\dfrac{1}{2}.$

By formula, a_n = ar^{n - 1}

$a_6 = \dfrac{3}{2}\Big(\dfrac{1}{2}\Big)^{6 - 1} = \dfrac{3}{2}\Big(\dfrac{1}{2}\Big)^5 = \dfrac{3}{2} \times \dfrac{1}{32} = \dfrac{3}{64}. \\[1em] a_n = \dfrac{3}{2}\Big(\dfrac{1}{2}\Big)^{n - 1} = \dfrac{3}{2 \times 2^n \times 2^{-1}} = \dfrac{3}{2^n}.$

Hence, the 6th term is $\dfrac{3}{64} \text{ and nth term is } \dfrac{3}{2^n}.$

Find the 6th term from the end of the list of numbers 3, -6, 12, -24, .... , 12288.

Answer

The given list of numbers 3, -6, 12, -24, .... , 12288 is a G.P. with last term = l = 12288 and the common ratio = r = -2.

By formula, nth term from end = $l\Big(\dfrac{1}{r}\Big)^{n - 1}$

$\text{7th term from end } = 12288\Big(-\dfrac{1}{2}\Big)^{6 - 1} \\[1em] = 12288\Big(-\dfrac{1}{2}\Big)^5 \\[1em] = -\dfrac{12288}{32} \\[1em] = -384.$

Hence, the 6th term from end of the G.P. is -384.

Which term of the G.P.

(i) 2, $2\sqrt{2}$, 4, .... is 128?

(ii) $1, \dfrac{1}{3}, \dfrac{1}{9}, .... \text{ is } \dfrac{1}{243}$ ?

Answer

(i) Given a = 2, r = $\sqrt{2}$.

Let nth term be 128.

By formula, a_n = ar^{n - 1}.

$\Rightarrow 128 = 2(\sqrt{2})^{n - 1} \\[1em] \Rightarrow \dfrac{128}{2} = (2)^{(n - 1)/2} \\[1em] \Rightarrow 64 = (2)^{(n - 1)/2} \\[1em] \Rightarrow 2^6 = (2)^{(n - 1)/2} \\[1em] \Rightarrow \dfrac{n - 1}{2} = 6 \\[1em] \Rightarrow n - 1 = 12 \\[1em] \Rightarrow n = 13.$

Hence, 128 is 13th term of the G.P.

(ii) Given a = 1, r = $\dfrac{1}{3}$.

Let nth term be $\dfrac{1}{243}$.

By formula, a_n = ar^{n - 1}.

$\Rightarrow \dfrac{1}{243} = 1\Big(\dfrac{1}{3}\Big)^{n - 1} \\[1em] \Rightarrow \dfrac{1}{3^5} = \dfrac{1}{3^{(n - 1)}} \\[1em] \Rightarrow n - 1 = 5 \\[1em] \Rightarrow n = 6.$

Hence, $\dfrac{1}{243}$ is 6th term of the G.P.

Determine the 12th term of a G.P. whose 8th term is 192 and common ratio is 2.

Answer

Given, a₈ = 192 and r = 2.

By formula, a_n = ar^{n - 1}.

⇒ a₈ = a(2)^{(8 - 1)}
⇒ 192 = a(2)⁷
⇒ a = $\dfrac{192}{2^7} = \dfrac{192}{128} = \dfrac{3}{2}.$

12th term of the G.P. is a₁₂,

$\Rightarrow a_{12} = \dfrac{3}{2}(2)^{12 - 1} \\[1em] \Rightarrow a_{12} = \dfrac{3}{2} \times 2^{11} \\[1em] \Rightarrow a_{12} = 3 \times 2^{10} \\[1em] \Rightarrow a_{12} = 3 \times 1024 \\[1em] \Rightarrow a_{12} = 3072.$

Hence, the 12th term of the G.P. is 3072.

In a G.P., the third term is 24 and 6th term is 192. Find the 10th term.

Answer

Given, a₃ = 24, a₆ = 192.

By formula, a_n = ar^{n - 1}.

⇒ a₃ = ar^{(3 - 1)}
⇒ ar² = 24.             (Eq 1)

⇒ a₆ = ar^{(6 - 1)}
⇒ ar⁵ = 192.            (Eq 2)

Dividing Eq 2 by Eq 1

$\Rightarrow \dfrac{ar^5}{ar^2} = \dfrac{192}{24} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = (2)^3 \\[1em] \Rightarrow r = 2$

Putting value of r in Eq 1,

$\Rightarrow a(2)^2 = 24 \\[1em] \Rightarrow 4a = 24 \\[1em] \Rightarrow a = 6 \\[1em] \text{10th term of G.P.} = a_{10} \\[1em] \Rightarrow a_{10} = 6(2)^{10 - 1} \\[1em] = 6(2)^9 = 6 \times 512 = 3072.$

Hence, the 10th term of the G.P. is 3072.

Find the number of terms of a G.P. whose first term is $\dfrac{3}{4},$ common ratio is 2 and the last term is 384.

Answer

Let the number of terms be n.

Given, a = $\dfrac{3}{4}$, r = 2 and a_n = 384.

By formula, a_n = ar^{n - 1}.

$\Rightarrow 384 = \dfrac{3}{4}(2)^{n - 1} \\[1em] \Rightarrow \dfrac{384 \times 4}{3} = (2)^{n - 1} \\[1em] \Rightarrow (2)^{n - 1} = 512 \\[1em] \Rightarrow (2)^{n - 1} = (2)^9 \\[1em] \Rightarrow n - 1 = 9 \\[1em] \Rightarrow n = 10.$

Hence, the number of terms in the G.P. are 10.

Find the value of x such that

(i) $-\dfrac{2}{7}, x, -\dfrac{7}{2}$ are three consecutive terms of a G.P.

(ii) x + 9, x - 6 and 4 are three consecutive terms of a G.P.

(iii) x, x + 3, x + 9 are first three terms of a G.P.

Answer

(i) Since, $-\dfrac{2}{7}, x, -\dfrac{7}{2}$ are three consecutive terms of a G.P.

So,

$\Rightarrow \dfrac{x}{-\dfrac{2}{7}} = r = \dfrac{-\dfrac{7}{2}}{x} \\[1em] \Rightarrow \dfrac{x}{-\dfrac{2}{7}} = \dfrac{-\dfrac{7}{2}}{x} \\[1em] \Rightarrow x^2 = -\dfrac{7}{2} \times -\dfrac{2}{7} \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x^2 - 1 = 0 \\[1em] \Rightarrow (x - 1)(x + 1) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x + 1 = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = -1.$

Hence, the value of x = 1 or -1.

(ii) Since, x + 9, x - 6 and 4 are three consecutive terms of a G.P.

So,

$\Rightarrow \dfrac{x - 6}{x + 9} = r = \dfrac{4}{x - 6} \\[1em] \Rightarrow \dfrac{x - 6}{x + 9} = \dfrac{4}{x - 6} \\[1em] \Rightarrow (x - 6)^2 = 4(x + 9) \\[1em] \Rightarrow x^2 + 36 - 12x = 4x + 36 \\[1em] \Rightarrow x^2 - 12x - 4x + 36 - 36 = 0 \\[1em] \Rightarrow x^2 - 16x = 0 \\[1em] \Rightarrow x(x - 16) = 0 \\[1em] \Rightarrow x = 0 \text{ or } x - 16 = 0 \\[1em] \Rightarrow x = 0 \text{ or } x = 16.$

Hence, the value of x = 0 or 16.

(iii) Since, x, x + 3 and x + 9 are first three terms of a G.P.

So,

$\Rightarrow \dfrac{x + 3}{x} = r = \dfrac{x + 9}{x + 3} \\[1em] \Rightarrow \dfrac{x + 3}{x} = \dfrac{x + 9}{x + 3} \\[1em] \Rightarrow (x + 3)^2 = x(x + 9) \\[1em] \Rightarrow x^2 + 9 + 6x = x^2 + 9x \\[1em] \Rightarrow x^2 - x^2 + 9 + 6x - 9x = 0 \\[1em] \Rightarrow 9 - 3x = 0 \\[1em] \Rightarrow 3x = 9 \\[1em] \Rightarrow x = 3. \\[1em]$

Hence, the value of x = 3.

If the fourth, seventh and tenth terms of a G.P. are x, y, z respectively, prove that x, y, z are in G.P.

Answer

Given, a₄ = x, a₇ = y and a₁₀ = z.

By formula, a_n = ar^{n - 1}.

⇒ x = a₄ = a(r)³

⇒ y = a₇ = a(r)⁶

⇒ z = a₁₀ = a(r)⁹

From above equations,

$\Rightarrow \dfrac{y}{x} = \dfrac{ar^6}{ar^3} = r^3. \\[1em] \Rightarrow \dfrac{z}{y} = \dfrac{ar^9}{ar^6} = r^3.$

The above equations prove that the common ratio between x, y and z is r³.

Hence, x, y and z are in G.P. with common ratio = r³.

The 5th, 8th and 11th terms of a G.P. are p, q and s respectively. Show that q² = ps.

Answer

Let the first term of the G.P. be a and common ratio = r.

Given,

⇒ a₅ = ar⁴ = p

⇒ a₈ = ar⁷ = q

⇒ a₁₁ = ar¹⁰ = s

We need to prove q² = ps.

L.H.S. = q²

⇒ q² = q × q = ar⁷ × ar⁷ = a²r¹⁴.

R.H.S. = ps

⇒ ps = p × s = ar⁴ × ar¹⁰ = a²r¹⁴.

∴ L.H.S. = R.H.S. = a²r¹⁴.

Hence, proved that q² = ps.

If a, a² + 2 and a³ + 10 are in G.P., then find the value(s) of a.

Answer

Since a, a² + 2 and a³ + 10 are in G.P.

$\therefore \dfrac{a^2 + 2}{a} = r = \dfrac{a^3 + 10}{a^2 + 2} \\[1em] \Rightarrow \dfrac{a^2 + 2}{a} = \dfrac{a^3 + 10}{a^2 + 2} \\[1em] \Rightarrow (a^2 + 2)^2 = a(a^3 + 10) \\[1em] \Rightarrow a^4 + 4 + 4a^2 = a^4 + 10a \\[1em] \Rightarrow a^4 - a^4 + 4a^2 - 10a + 4 = 0 \\[1em] \Rightarrow 4a^2 - 10a + 4 = 0 \\[1em] \Rightarrow 4a^2 - 8a - 2a + 4 = 0 \\[1em] \Rightarrow 4a(a - 2) - 2(a - 2) = 0 \\[1em] \Rightarrow (4a - 2)(a - 2) = 0 \\[1em] \Rightarrow 4a - 2 = 0 \text{ or } a - 2 = 0 \\[1em] \Rightarrow a = \dfrac{2}{4} \text{ or } a = 2. \\[1em] \Rightarrow a = \dfrac{1}{2} \text{ or } a = 2.$

Hence, the required value(s) of a are $\dfrac{1}{2}$ and 2.

Find the geometric progression whose 4th term is 54 and the 7th term is 1458.

Answer

Given, a₄ = 54 and a₇ = 1458.

By formula, a_n = ar^{n - 1}.

⇒ a₄ = a(r)^{4 - 1}
⇒ 54 = ar³     (Eq 1)

⇒ a₇ = a(r)^{7 - 1}
⇒ 1458 = a(r)⁶     (Eq 2)

Dividing Eq 2 by Eq 1,

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{1458}{54} \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r = \sqrt[3]{27} \\[1em] \Rightarrow r = 3$

Putting value of r in Eq 1,

$\Rightarrow a(3)^3 = 54 \\[1em] \Rightarrow 27a = 54 \\[1em] \Rightarrow a = 2.$

a₂ = ar = 2 × 3 = 6
a₃ = ar² = 2(3)² = 2 × 9 = 18
a₄ = ar³ = 2(3)³ = 2 × 27 = 54.

Hence, the required G.P. is 2, 6, 18, 54, ...

The sum of first three terms of a G.P. is $\dfrac{39}{10}$ and their product is 1. Find the common ratio and the terms.

Answer

Given, S₃ = $\dfrac{39}{10}$ and a₁ × a₂ × a₃ = 1.

Let the three numbers that are in G.P. be $\dfrac{a}{r}, a, ar.$

$\therefore \dfrac{a}{r} \times a \times ar = 1 \\[1em] \Rightarrow a^3 = 1 \\[1em] \Rightarrow a = 1. \\[1em] S_3 = \dfrac{39}{10} \\[1em] \Rightarrow \dfrac{a}{r} + a + ar = \dfrac{39}{10} \\[1em] \Rightarrow a\Big(\dfrac{1}{r} + 1 + r\Big) = \dfrac{39}{10} \\[1em] \Rightarrow 1\Big(\dfrac{1 + r + r^2}{r}\Big) = \dfrac{39}{10} \\[1em] \Rightarrow 10(1 + r + r^2) = 39r \\[1em] \Rightarrow 10 + 10r + 10r^2 = 39r \\[1em] \Rightarrow 10r^2 + 10r - 39r + 10 = 0 \\[1em] \Rightarrow 10r^2 - 29r + 10 = 0 \\[1em] \Rightarrow 10r^2 - 25r - 4r + 10 = 0 \\[1em] \Rightarrow 5r(2r - 5) - 2(2r - 5) = 0 \\[1em] \Rightarrow (5r - 2)(2r - 5) = 0 \\[1em] \Rightarrow 5r - 2 = 0 \text{ or } 2r - 5 = 0 \\[1em] \Rightarrow r = \dfrac{2}{5} \text{ or } r = \dfrac{5}{2} \\[1em] \text{First taking r } = \dfrac{2}{5} \\[1em] \therefore \text{Terms are : } \dfrac{a}{r} = \dfrac{1}{\dfrac{2}{5}} = \dfrac{5}{2}, a = 1, ar = 1 \times \dfrac{2}{5} = \dfrac{2}{5}. \\[1em] \text{Now, taking r } = \dfrac{5}{2} \\[1em] \therefore \text{Terms are : } \dfrac{a}{r} = \dfrac{1}{\dfrac{5}{2}} = \dfrac{2}{5}, a = 1, ar = 1 \times \dfrac{5}{2} = \dfrac{5}{2}. \\[1em]$

Hence, common ratio is $\dfrac{2}{5}$ or $\dfrac{5}{2}$ and terms are $\dfrac{5}{2}, 1, \dfrac{2}{5}$ or $\dfrac{2}{5}, 1, \dfrac{5}{2}.$

Three numbers are in A.P. and their sum is 15. If 1, 4 and 19 are added to these numbers respectively, the resulting numbers are in G.P. Find the numbers.

Answer

Let three numbers that are in A.P. be a - d, a, a + d.

Given, sum of three numbers = 15.

⇒ a - d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5.

By adding 1, 4 and 19 in the terms become,

⇒ a - d + 1, a + 4 and a + d + 19
⇒ 5 - d + 1, 5 + 4 and 5 + d + 19
⇒ 6 - d, 9 and d + 24.

According to question these terms become in G.P.,

$\therefore \dfrac{9}{6 - d} = \dfrac{d + 24}{9} \\[1em] \Rightarrow (d + 24)(6 - d) = 81 \\[1em] \Rightarrow 6d - d^2 + 144 - 24d = 81 \\[1em] \Rightarrow 6d - d^2 - 24d + 144 - 81 = 0 \\[1em] \Rightarrow -18d - d^2 + 63 = 0 \\[1em] \Rightarrow d^2 + 18d - 63 = 0 \\[1em] \Rightarrow d^2 + 21d - 3d - 63 = 0 \\[1em] \Rightarrow d(d + 21) - 3(d + 21) = 0 \\[1em] \Rightarrow (d - 3)(d + 21) = 0 \\[0.5em] \Rightarrow d - 3 = 0 \text{ or } d + 21 = 0 \\[1em] \Rightarrow d = 3 \text{ or } d = -21.$

Taking d = 3,

∴ a - d = 5 - 3 = 2, a = 5, a + d = 5 + 3 = 8.

Taking d = - 21,

∴ a - d = 5 -(-) 21 = 26, a = 5, a + d = 5 +(-) 21 = - 16.

Hence, the required numbers are 2, 5 and 8 or 26, 5 and -16.