Arun Sharma # progression

Question

In an infinite geometric progression ,each term is equal to $2$ times the sum of the terms that follow . If the first term of the series is $8$ ,find the sum of the series?

Answer 1

First Term is 8, So the Sum of All the terms that follow it will be 4. Hence, Total Sum = $12$

Comments

u mean 4 is sum of some terms of GP

what those terms could be?

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yes, hard to find GP

12 may be ans :)

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Can u. write the series?

No. The Data given in the Question might not be sufficient for writing the whole series. Even no need to think of it as a GP series, Just think of some sequence in which the given condition "Each term is 2 times the sum of all the terms that follow" holds. The Given data is sufficient to get answer for what has been asked. That's all needed.  

Answer 2

1st point is : 

it is an infinite GP Series.

2nd point :

each term is equal to 2 times the sum of the terms that follow.

Now, the first term is $8$.

∴ The series must be decreasing & $8 = 4 \times 2$

∴ Sum of the terms that follow  $8$ is $4$.

∴ $\color{green}{\text{Sum of the series}}$ = $\text{First term  +  Sum of the terms that follow the first term}$

$ \qquad \qquad = 8 + 4$

$ \qquad \qquad = \color{Gold}{12}$

Comments

@all, please correct me where I am wrong..

Let infinite GP with common ratio 'r' is :- a,ar,ar² ,ar³, ar⁴,......

since , each term is equal to 2 times the sum of the terms that follow it ..

So ,a = 2(ar+ar² + ar³+........) ---------equation (1)

Since , a + ar + ar²  + ar³ + .... = a/(1-r) , r<1

So , ar + ar²  + ar³ + .... = a/(1-r) - a

So, according to equation (1),

a= 2[(a/(1-r))-a]...after solving , r = 1/3..

So,now infinite GP is :- 8,8/3,8/9,....

Sum of this infinite series is :- 8/(1-(1/3)) = 12..

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And We have a Winner.

Add this as an Answer to the Question. It is slightly more time taking approach for this simple question But knowing such different approaches is GOOD for knowledge.

You can also justify the $\left | r \right | \leq1$ By Saying that Since it is a Decreasing GP, $\left | r \right | \leq1$ will hold.

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yes, really going awesome discussion :)