a convergent geometric sequence with first term a is such that the sum of the terms after the n^(th) term is three times the n^(th) term, find the common ratio and show that its sum to infinity is 4a.


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Question Number 92820 by prince 5 last updated on 13/May/20

$a c o n v e r g e n t g e o m e t r i c s e q u e n c e w i t h$ $f i r s t t e r m a i s s u c h t h a t t h e s u m o f$ $t h e t e r m s a f t e r t h e n^{t h} t e r m i s$ $t h r e e t i m e s t h e n^{t h} t e r m, f i n d t h e$ $c o m m o n r a t i o a n d s h o w t h a t i t s$ $s u m t o i n f i n i t y i s 4 a .$
	Answered by prakash jain last updated on 12/May/20

	$For a geometric seres$ $fisrt term a$ $common ratio r$ $n^{t h} term = {a r}^{n - 1}$ $sum = \frac{a}{1 - r} (∣ r ∣< 1 for covergence)$ $sum after n^{t h} term is same as$ $geometric series with first term$ $as (n + 1) th term and common ratio r .$ $\frac{{a r}^{n}}{1 - r} = 3 {a r}^{n - 1}$ $\frac{r}{1 - r} = 3 \Rightarrow r = 3 - 3 r \Rightarrow r = \frac{3}{4}$ $Sum of series = \frac{a}{1 - r} = 4 a$
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