Question

The sum of an infinite geometric series of real numbers is 14, and the sum of the cubes of the terms of this series is 392. then the first term of the series is

Answer 1

let the series be
$a \: {a}^{} r \: a {r}^{2} .......$
sum
$s = \frac{a}{1 - r} = 14.....(1)$
when cubed, series become
${a}^{3} \: a {}^{3} {r}^{3} .......$
sum=
$\frac{ {a}^{3} }{1 - {r}^{3} } = 392....(2)$
solve these two eqns

Answer 2

Given :

The sum of an infinite geometric series of real numbers is 14, and the sum of the cubes of the terms of this series is 392.

To Find:

The first term of the series is

Solution:

We are given that The sum of an infinite geometric series of real numbers is 14.

Formula of sum of infinite terms of GP :

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

So, $\frac{a}{1-r}=14 ------1$

We are also given that the sum of the cubes of the terms of this series is 392.

So, $\frac{a^3}{1-r^3}=392 ---- 2$

Cubing equation 1 and Divide 1 and 2

So,$\frac{14^3}{392}=\frac{(\frac{a}{1-r})^3}{\frac{a^3}{1-r^3}}$

$\frac{14^3}{392}=\frac{1-r^3}{(1-r)^3}\\7(1-r)^2=1+r^2-r\\6r^2-13r+6=0\\r=\frac{2}{3},\frac{3}{2}$

For$r =\frac{2}{3}$

$\frac{a}{1-\frac{2}{3}}=14\\a=7$

For$r = \frac{3}{2}$

$\frac{a}{1-\frac{3}{2}}=14$

a=-7

So,the first term of the series is 7 or -7