The sum of the first two terms of a geometric sequence is $\frac{8}{9}$ $8/9$ and of the first three terms is $\frac{26}{27}$ $26/27$ . What is the first term, common ratio and sum to infinity ?

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Answer 1 · 2016-08-14T17:12:40

The first term can be $\frac{2}{3}$ $2/3$ with common ratio $\frac{1}{3}$ $1/3$ and sum $1$ $1$ .

The first term can be $\frac{32}{27}$ $32/27$ with common ratio $- \frac{1}{4}$ $-1/4$ and sum $\frac{128}{135}$ $128/135$

These are the only two possibilities.

If the first term is $a$ $a$ and the common ratio is $r$ $r$ , then we are given:

$a (1 + r) = \frac{8}{9}$ $a(1+r) = 8/9$

$a (1 + r + r^{2}) = \frac{26}{27}$ $a(1+r+r^2) = 26/27$

So:

$13/12 = (26*9)/(27*8) = (color(red)(cancel(color(black)(a)))(1+r+r^2))/(color(red)(cancel(color(black)(a)))(1+r)) = (1+r+r^2)/(1+r)$

Cross multiply to get:

$13(1+r) = 12(1+r+r^2)$

Rearrange to get:

$0 = 12r^2-r-1 = (3r-1)(4r+1)$

So:

$r = 1/3 " "$ or $" " r = -1/4$

If $r=1/3$ then $a=8/(9(1+r)) = 2/3$

The sum of the whole series is then $(2/3) / (1-1/3) = 1$

If $r=-1/4$ then $a=8/(9(1+r)) = 32/27$

The sum of the whole series is then $(32/27) / (1-(-1/4)) = 128/135$

The sum of the first two terms of a geometric sequence is 8/9898/9 and of the first three terms is 26/27262726/27. What is the first term, common ratio and sum to infinity ?