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  • 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

 

Answers (1)

best_answer

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.

Posted by

prateek

View full answer

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