Let a_n be a G.P.

Question

Let $\{a_n\}$ be a G.P. of increasing positive terms such that $a_2 \cdot a_3 \cdot a_4 = 64$ and $a_1+a_3+a_5 = \dfrac{813}{7}$. Then $a_3+a_5+a_7$ is equal to:

Accepted Answer

$a_2a_3a_4 = \dfrac{a}{r}\cdot a\cdot ar = a^3 = 64 \Rightarrow a = 4$.

$$a_1+a_3+a_5 = \frac{a}{r^2}+a+ar^2 = \frac{4}{r^2}+4+4r^2 = \frac{813}{7}.$$

Let $t=r^2$: $28t^2 - 785t + 28 = 0 \Rightarrow (28t-1)(t-28)=0 \Rightarrow t = 28$ or $t=\tfrac{1}{28}$. Since the G.P. is increasing, $r^2 = 28$.

$$a_3+a_5+a_7 = a r^2 + ar^4 + ar^6 = r^2\left(a+\frac{a}{r^2}\cdot r^4\right)\ \text{...} = r^2(a_1+a_3+a_5) = 28\times\frac{813}{7} = 3252.$$

Solutions · 1