The sum of the first two terms of a geometric sequence is #8/9# and of the first three terms is #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 #2/3# with common ratio #1/3# and sum #1#.

The first term can be #32/27# with common ratio #-1/4# and sum #128/135#

These are the only two possibilities.

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

#a(1+r) = 8/9#

#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#