Class 11: Geometric Progression – Exercise 20.6 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note:

1. If there are $n$ Geometric Means inserted between $a$ and $b$ , then

$r = \Big($ $\frac{b}{a}$ $\Big)^{\frac{1}{n+1}}$

2. Geometric mean between $a$ and $b$ is $\sqrt{ab}$

Question 1: Insert 6 geometric means between $27$ and $\frac{1}{81}$ .

Answer:

Let $G_1, G_2, G_3, G_4, G_5, G_6$ be the $6$ GMs inserted between $a = 27$ and $b =$ $\frac{1}{81}$ .

$n = 6$ ( As $6$ geometric means are being inserted).

$\therefore r = \Big($ $\frac{1/81}{27}$ $\Big)^{\frac{1}{6+1}}$ $= \Big($ $\frac{1}{3^7}$ $\Big)^{\frac{1}{7}}$ $=$ $\frac{1}{3}$ . Therefore,

$G_1 = ar^{ } = 27 \Big($ $\frac{1}{3}$ $\Big)^{ } = 9$

$G_2 = ar^{2 } = 27 \Big($ $\frac{1}{3}$ $\Big)^{2 } = 3$

$G_3 = ar^{3 } = 27 \Big($ $\frac{1}{3}$ $\Big)^{3 } = 1$

$G_4 = ar^{4 } = 27 \Big($ $\frac{1}{3}$ $\Big)^{4 } =$ $\frac{1}{3}$

$G_5 = ar^{5 } = 27 \Big($ $\frac{1}{3}$ $\Big)^{5 } =$ $\frac{1}{9}$

$G_6 = ar^{6 } = 27 \Big($ $\frac{1}{3}$ $\Big)^{ 6} =$ $\frac{1}{27}$

Hence the 6 geometric means between $a = 27$ and $b =$ $\frac{1}{81}$ are $9, 3, 1,$ $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}$

$\\$

Question 2: Insert 5 geometric means between $16$ and $\frac{1}{4}$ .

Answer:

Let $G_1, G_2, G_3, G_4, G_5$ be the $5$ GMs inserted between $a = 16$ and $b =$ $\frac{1}{4}$ .

$n = 5$ ( As $5$ geometric means are being inserted).

$\therefore r = \Big($ $\frac{1/4}{16}$ $\Big)^{\frac{1}{5+1}}$ $= \Big($ $\frac{1}{3^6}$ $\Big)^{\frac{1}{6}}$ $=$ $\frac{1}{2}$ . Therefore,

$G_1 = ar^{ } = 16 \Big($ $\frac{1}{2}$ $\Big)^{ } = 8$

$G_2 = ar^{2 } = 16 \Big($ $\frac{1}{2}$ $\Big)^{2 } = 4$

$G_3 = ar^{3 } = 16 \Big($ $\frac{1}{2}$ $\Big)^{3 } = 2$

$G_4 = ar^{4 } = 16 \Big($ $\frac{1}{2}$ $\Big)^{4 } = 1$

$G_5 = ar^{5 } = 16 \Big($ $\frac{1}{2}$ $\Big)^{5 } =$ $\frac{1}{2}$

Hence the 5 geometric means between $a = 16$ and $b =$ $\frac{1}{4}$ are $8, 4, 2, 1,$ $\frac{1}{2}$

$\\$

Question 3: Insert 5 geometric means between $\frac{32}{9}$ and $\frac{81}{2}$ .

Answer:

Let $G_1, G_2, G_3, G_4, G_5$ be the $5$ GMs inserted between $\frac{32}{9}$ and $\frac{81}{2}$ .

$n = 5$ ( As $5$ geometric means are being inserted).

$\therefore r = \Big($ $\frac{81/2}{32/9}$ $\Big)^{\frac{1}{5+1}}$ $= \Big($ $\frac{3^6}{2^6}$ $\Big)^{\frac{1}{6}}$ $=$ $\frac{3}{2}$ . Therefore,

$G_1 = ar^{ } = \Big($ $\frac{32}{9}$ $\Big) \Big($ $\frac{3}{2}$ $\Big)^{ } =$ $\frac{16}{3}$

$G_2 = ar^{2 } = \Big($ $\frac{32}{9}$ $\Big) \Big($ $\frac{3}{2}$ $\Big)^{2 } = 8$

$G_3 = ar^{3 } = \Big($ $\frac{32}{9}$ $\Big) \Big($ $\frac{3}{2}$ $\Big)^{3 } = 12$

$G_4 = ar^{4 } = \Big($ $\frac{32}{9}$ $\Big) \Big($ $\frac{3}{2}$ $\Big)^{4 } = 18$

$G_5 = ar^{5 } = \Big($ $\frac{32}{9}$ $\Big) \Big($ $\frac{3}{2}$ $\Big)^{5 } = 27$

Hence the 5 geometric means between $\frac{32}{9}$ and $\frac{81}{2}$ are $\frac{16}{3}$ $8, 12, 18, 27$

$\\$

Question 4: Find the geometric means of the following pairs of numbers:

i) $2 \text{ and } 8$ ii) $a^3b \text{ and } ab^3$ iii) $-8 \text{ and } - 2$

Answer:

i) $2 \text{ and } 8$

Geometric mean between $2 \text{ and } 8 = \sqrt{2 \times 8} = \sqrt{16} = 4$

ii) $a^3b \text{ and } ab^3$

Geometric mean between $a^3b \text{ and } ab^3 = \sqrt{a^3b \times ab^3} = \sqrt{a^4b^4} = a^2b^2$

iii) $-8 \text{ and } - 2$

Geometric mean between $-8 \text{ and } -2 = \sqrt{(-8) \times (-2)} = \sqrt{16} = 4$

$\\$

Question 5: If $a$ is the G.M. of $2$ and $\frac{1}{4}$ , find $a$

Answer:

$a =$ $\sqrt{2 \times \frac{1}{4} }$ $=$ $\frac{1}{\sqrt{2}}$

$\\$

Question 6: Find the two numbers whose A.M. is $25$ and GM is $20$ .

Answer:

Given A.M. $= 25 \Rightarrow$ $\frac{a+b}{2}$ $= 25 = a+b = 50$ … … … … … i)

Also GM $= 20 \Rightarrow \sqrt{ab} = 20 \Rightarrow ab = 400$ … … … … … ii)

Solving i) and ii) we get

$a( 50 -a ) = 400$

$\Rightarrow a^2 - 50a + 400 = 0$

$\Rightarrow ( a - 40)(a - 10) = 0$

$\Rightarrow a = 40$ or $a = 10$

When $a = 40, b =$ $\frac{400}{40}$ $= 10$

When $a = 10, b =$ $\frac{400}{10}$ $= 40$

Hence the two numbers are $10$ and $40$

$\\$

Question 7: Construct a quadratic in $x$ such that A.M. of its roots is $A$ and G.M. is $G$ .

Answer:

Let the two roots of the quadratic equation be $a$ and $b$

$\therefore A =$ $\frac{a+b}{2}$ $\Rightarrow a + b = 2 A$ … … … … … i)

Also $G^2 = ab$ … … … … … ii)

We know the quadratic equation having roots $a$ and $b$ is given by $x^2 - ( a+b) x + ab = 0$

$\Rightarrow x^2 - 2A x + G^2= 0$

$\\$

Question 8: The sum of two numbers is $6$ times their geometric mean, show that the numbers are in the ratio $(3 + 2\sqrt{2}): ( 3 - 2\sqrt{2})$ .

Answer:

Let the two numbers be $a$ and $b$

Given $a+b = 6 \sqrt{ab}$

$\Rightarrow$ $\frac{a+b}{2\sqrt{ab}} = \frac{3}{1}$

Applying componendo and dividendo

$\frac{a+b + 2\sqrt{ab}}{a+b - 2\sqrt{ab}} = \frac{3+1}{3-1}$

$\Rightarrow$ $\Big( \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} \Big)^2 = \frac{4}{2}$

$\Rightarrow$ $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \frac{\sqrt{2}}{1}$

Applying componendo and dividendo once again

$\frac{\sqrt{a}+\sqrt{b}+ \sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b} - \sqrt{a} + \sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1}$

$\Rightarrow$ $\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1}$

$\Rightarrow$ $\frac{a}{b} =\Big( \frac{\sqrt{2}+1}{\sqrt{2}-1}\Big)^2$

$\Rightarrow$ $\frac{a}{b} =\frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}}$

$\Rightarrow$ $\frac{a}{b} =\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$ . Hence proved.

$\\$

Question 9: If AM and GM of roots of a quadratic equation are $8$ and $5$ respectively, then obtain the quadratic equation.

Answer:

Let the two roots of the quadratic equation be $a$ and $b$

Given AM $= 8 \Rightarrow$ $\frac{a+b}{2}$ $= 8 \Rightarrow a + b = 16$ … … … … … i)

Also GM $= 5 \Rightarrow \sqrt{ab} = 5 \Rightarrow ab = 25$

Therefore the quadratic equation will be $x^2 - ( a+b) x + ab = 0$

$\Rightarrow x^2 - 16x + 25 = 0$

$\\$

Question 10: If AM and GM of two positive numbers $a \ and \ b$ are $10 \ and \ 8$ respectively, find the numbers.

Answer:

Given AM $= 10 \Rightarrow$ $\frac{a+b}{2}$ $=10 \Rightarrow a + b = 20$ … … … … … i)

Also GM $= 8 \Rightarrow \sqrt{ab} = 8 \Rightarrow ab = 64$ … … … … … ii)

Solving i) and ii) we get

$a( 20-a) = 6$

$\Rightarrow a^2 - 20a + 64 = 0$

$\Rightarrow ( a-16)(a-4) = 0$

$\Rightarrow a = 16 \ or \ a = 4$

When $a = 16, b =$ $\frac{64}{16}$ $= 4$

When $a = 4, b =$ $\frac{64}{4}$ $= 16$

Hence the two numbers are $4$ and $16$ .

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Question 11: Prove that the product of $n$ geometric means between two quantities is equal to the $n^{th}$ power of a geometric mean of those two quantities.

Answer:

Let $G_1, G_2, G_3, \ldots , G_n$ be the $n$ GMs inserted between $a$ and $b$

Let $r$ be the common ratio.

$\therefore r = \Big($ $\frac{b}{a}$ $\Big)^{\frac{1}{n+1}}$

$G_1 = ar$

$G_2 = ar^2$

$\ldots$

$G_n = ar^n$

Let $G$ be the GM between $a$ and $b$

$\therefore G = \sqrt{ab} \Rightarrow G^2 = ab$

Now Given $G_1 \times G_2 \times \ldots \times G_n$

$= ar \times ar^2 \times \ldots \times ar^n$

$= a^n \cdot r^{1+2+3+\ldots+ n}$

$= a^n \Big[ \Big( \frac{b}{a} \Big)^{\frac{1}{n+1}} \Big]^{\frac{n(n+10)}{2}}$

$= a^n \Big( \frac{b}{a} \Big)^{\frac{n}{2}}$

$= a^{\frac{n}{2}} b^{\frac{b}{2}}$

$= ( \sqrt{ab})^n$

$= (G)^n$ . Hence proved.

$\\$

Question 12: If the A.M. of two positive numbers $a$ and $b (a>b)$ is twice their geometric mean. Prove that: $a:b = ( 2 + \sqrt{3}):(2 - \sqrt{3})$

Answer:

Let the two numbers be $a$ and $b$

Given A.M. = 2 G.M.

Given $\frac{a+b}{2} = 2 \sqrt{ab}$

$\Rightarrow$ $\frac{a+b}{2\sqrt{ab}} = \frac{2}{1}$

Applying componendo and dividendo

$\frac{a+b + 2\sqrt{ab}}{a+b - 2\sqrt{ab}} = \frac{2+1}{2-1}$

$\Rightarrow$ $\Big( \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} \Big)^2 = \frac{3}{1}$

$\Rightarrow$ $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \frac{\sqrt{3}}{1}$

Applying componendo and dividendo once again

$\frac{\sqrt{a}+\sqrt{b}+ \sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b} - \sqrt{a} + \sqrt{b}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$

$\Rightarrow$ $\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$

$\Rightarrow$ $\frac{a}{b} =\Big( \frac{\sqrt{3}+1}{\sqrt{3}-1}\Big)^2$

$\Rightarrow$ $\frac{a}{b} =\frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}}$

$\Rightarrow$ $\frac{a}{b} =\frac{2+\sqrt{3}}{2-\sqrt{3}}$ . Hence proved.

$\\$

Question 13: If one A.M., $A$ and two geometric means $G_1$ and $G_2$ inserted between any two positive numbers, show that $\frac{{G_1}^2}{G_2}+\frac{{G_2}^2}{G_1}$ $= 2A$