The sum of the 3rd and the 4th terms of a G.P.

Question

The sum of the 3rd and the 4th terms of a G.P. is 60, and the product of its first three terms is 1000. If the first term of this G.P. is positive, then its 7th term is:

Accepted Answer

Let $a, ar, ar^2$ be the first three terms. Then $a(ar)(ar^2)=(ar)^3=1000\Rightarrow ar=10$.

Also $ar^2+ar^3=60\Rightarrow ar(r+r^2)=60\Rightarrow 10(r+r^2)=60\Rightarrow r^2+r-6=0\Rightarrow r=2,-3$.

For $r=2$: $a=5$ (positive, accepted); for $r=-3$: $a=-10/3$ (rejected, not positive).

$$T_7=ar^6=5(2)^6=5\times64=320.$$

Solutions · 1