# A geometric progression has ﬁrst term 1 and common ratio r. A second geometric progression has ﬁrst term 4 and common ratio (1/4)rThe two progressions have the same sum to inﬁnity, S.Find the...

A geometric progression has ﬁrst term 1 and common ratio r. A second geometric progression has ﬁrst term 4 and common ratio (1/4)r

The two progressions have the same sum to inﬁnity, S.

Find the values of r and S.

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The sum of a geometric progression is given by;

`S_n = (a(1-r^n))/(1-r)`

Where a is initial term r is common ratio and n is number of terms.

When n tends to `oo` the sum is given by;

`lim_(nrarroo) Sn = a/(1-r)`

For series `(1) rarrS_n = 1/(1-r)` --------------(1)

For series `(2)rarrS_n = 4/(1-(r/4))` ------------(2)

(1) = (2)

`1/(1-r) = 4/1-(r/4)`

`4-r = 16(1-r)`

`r = 12/15 `

`r = 4/5`

From (1)

`S = 1/(1-4/5) = 5`

*So the answers are r = 4/5 and S = 5*

`1^(st) G.P` a=1 ,common ratio =r

`S_(oo)=1/(1-r)` (i)

2nd G.P a=4 and comon ratio =r/4

S_(oo)= 4/(1-r/4)

=16/(4-r) (ii)

Given

1/(1-r)=16/(4-r)

4-r=16(1-r)

4=16-16r+r

4-16=-15r

-12=-15r

r=(-12)/(-15)

**r=4/5**

**S_(oo)=1/(1-4/5)=1/(1/5)**

**=5**

**Ans.**