Question

Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the `4^(t h)`by 18.

Answer 1

Let the required numbers be `a, ar, ar^(2)` and `ar^(3)`. Then,
`T_(3)-T_(1)=9rArr ar^(2)-a=9rArr a(r^(2)-1)=9` ...(i)
and `T_(2)-T_(4)=18 rArr ar-ar^(3)=18 rArr ar(1-r^(2))=18` ...(ii)
On dividing (ii) by (i), we get `r=-2`.
Putting `r=-2` in (i), we get `a(4-1)=9 rArr a=3`.
Hence, the required numbers are 3, -6, 12 and -24.