The sum of the first four terms of a geometric progression (G.P.) is 6512, and the sum of …

Question

The sum of the first four terms of a geometric progression (G.P.) is $\dfrac{65}{12}$, and the sum of their respective reciprocals is $\dfrac{65}{18}$. If the product of the first three terms of the G.P. is $1$, and the third term is $\alpha$, then $6\alpha$ is equal to:

Answer: 9

Accepted Answer

Let terms $a,ar,ar^2,ar^3$. $a(1+r+r^2+r^3)=\dfrac{65}{12}$, and $\dfrac{1+r+r^2+r^3}{ar^3}=\dfrac{65}{18}$.

Dividing: $a^2r^3 = \dfrac{18}{12}=\dfrac32$.

Product of first three terms: $a^3r^3=1$. Combined with $a^2r^3=\tfrac32$: dividing, $a = \dfrac23$, and $r^3 = \dfrac{3/2}{4/9}=\dfrac{27}{8} \Rightarrow r=\dfrac32$.

$$\alpha = ar^2 = \frac23\times\frac94 = \frac32 \implies 6\alpha = 9.$$

Solutions · 1