Question

In the geometric progression, the first is a and the common ratio is r. The sun of the first two terms is 12 and the third term is 16
Determine the ratio (ar^2)/(a+ar)
If the first term is larger than the second term, find the value of r.

Answer 1

a + ar = 12
a(1+r) = 12
a = 12/(1+r)

ar^2 = 16
a = 16/r^2

then 16/r^2 = 12/(1+r)
12r^2 = 16+16r
12r^2 - 16r - 16 = 0
3r^2 - 4r - 4 = 0
(r-2)(3r+2) = 0
r = 2 or r = -2/3

if r = 2, a = 16/4 = 4 ---> ar = 8, but t1 should be > t2, so, no good

if r = -2/3 , a = 16/(4/9) = 36
so first term is 36, 2nd term is -24
Ok then!

<b>r = -2/3

ar^2/(a+ar)
= 36(4/9) / (36-24)
= 4/3</b>

btw, for the case of r=2, a = 4
ar^2/(a+ar)
= 4(4)/(4+8)
= 16/12
= 4/3 , so that ratio is the same