Q

# Please help! The sum of an infinite geometric series of real number is 14, and the sum of the cube of the terms of this series is 392 then the first term of the series is

The sum of an infinite geometric series of real number is 14, and the sum of the cube of the terms of this series is 392 then the first term of the series is

• Option 1)

-14

• Option 2)

10

• Option 3)

7

• Option 4)

-5

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As learnt in

Sum of infinite terms of a GP -

$a+ar+ar^{2}+- - - - -= \frac{a}{1-r}\\here \left | r \right |<1$

- wherein

$a\rightarrow$ first term

$r\rightarrow$ common ratio

$S_{\infty}=14= \frac{a}{1-r}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(i)$

Also if series is $a^{3},a^{3}r^{3}, a^{3}r^{6}..........\infty$

$S_{\infty'}=392= \frac{a^{3}}{1-r^{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(ii)$

Divide $(i)$ and $(ii)$

$\frac{S^{3}_{\infty}}{S_{\infty'}}=\frac{14^{3}}{392}=\frac{\frac{a^{3}}{(1-r)^{3}}}{\frac{a^{3}}{1-r^{3}}}$

$\frac{196 \times 14}{392}=\frac{1-r^{3}}{(1-r)^{3}}=\frac{1-r+r^{2}}{(1-r)^2}$

$7(1-r)^{2}=1+r^{2}-r$

$7r^{2}+7-14r=r^{2}+1-r$

$6r^{2}-13r +6=0$

$6r^{2}-9r -4r+6=0\:\:\:\:\:\:\:\:\:\:\:\:r=\frac{2}{3},\frac{3}{2}$

For $r=\frac{2}{3}\:\:\:\:\:\:\:\:\:\:\:\:a=7$

Option 1)

-14

This option is incorrect.

Option 2)

10

This option is incorrect.

Option 3)

7

This option is correct.

Option 4)

-5

This option is incorrect.

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