Name: Arithmetic Mean
Uploaded: 2019-07-19
Duration: 3 min 49 s
Description: Given two numbers, a and b, we can insert a number A between them so that a, A, b is an A.P. Such a number A is called the arithmetic mean (A.M.) of a and b....

Question

The sum of an infinite G.P. is

5

and the sum of the squares of these terms is

15

find the G.P.

Accepted Answer

Let $' a'$ be the first term and $' r'$ be the common ratio of the infinite geometric progression.

Then the geometric progression is $a, a r, a r^{2}, a r^{3}, . . .$

The sum to infinity of this G.P. is given by

$S_{\infty} = \frac{a}{1 - r}, |r| < 1$

$\Rightarrow \frac{a}{1 - r} = 5 . . . (1)$

The sum of the squares of the terms of this G.P. is $a^{2} + a^{2} r^{2} + a^{2} r^{4} + a^{2} r^{6}$

This is also an infinite geometric progression with the first term as $' a^{2}'$ and common ratio as $' r^{2}'$

and sum to infinity of this G.P. is $15$ .

$5$

From $(1), we get$

$a = 5 (1 - r)$

On squaring both sides

$\Rightarrow a^{2} = 25 {(1 - r)}^{2}$

By equation $(2)$

$\Rightarrow 15 (1 - r^{2}) = 25 {(1 - r)}^{2}$

$\Rightarrow 15 (1 - r) (1 + r) = 25 {(1 - r)}^{2}$

$\Rightarrow 3 (1 + r) = 5 (1 - r)$ $[∵ r \neq 1]$

$\Rightarrow 3 + 3 r = 5 - 5 r$

$\Rightarrow 8 r = 2$

$∴ r = \frac{1}{4}$

Substitute the value of $r$ in equation $(1), \frac{a}{1 - (\frac{1}{4})} = 5$

$\Rightarrow \frac{4 a}{3} = 5$

$∵ a = \frac{15}{4}$

If $r = \frac{1}{4}$ and $a = \frac{15}{4}$ then the required geometric progression is $\frac{15}{4}, \frac{15}{16}, \frac{15}{64}, . . . . .$

The sum of an infinite G.P. is $5$ and the sum of the squares of these terms is $15$ find the G.P.

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