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If first terms of a decreasing infinite GP is 1 and sum is S, then sum of squares of its terms is??

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9 years ago

```							sum of infinite term in  G.P is a/(1-r). (where a is 1st term, and r is the ratio)
so here in this problem, 1/(1-r)=S    (Since a=1)
solving it u will get; r =1 - 1/s
since terms of a G.P are given by a, ar ,ar^2,ar^3..........and so on
so here u will get the following series...
i.e 1 + (1- 1/s) + (1-1/s)^2 + (1-1/s)^3+..........
on squaring each term u will get,
1^2 + (1- 1/s)^2 + (1 - 1/s)^4 +(1-1/s)^6.........[r =(1-1/s)^2] in this case                                     (1)
so the required sum of this series  is
s'=1 /(1 -r) substitute value of r from (1)
finally u will get s' =s^2/(2s-1) ans..   .hope my answer will satisfy u.
```
9 years ago
```							let the GP is a + ar + ar2 + ar3 ...............infinity
sum of infinite GP S = a/1-r
S = 1/1-r                     (a=1)
r = S-1/S       ..........1
now gp formed after squaring is a2 + a2r2 + a2r4  ...............infinity
its common ratio is r2
now sum is S1 = a/1-r2 =1/(1-r)(1+r)
from eq 1 putting value of r
S1 = S2/(2S-1)
```
9 years ago
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