Let A_1, A_2, A_3, be an increasing geometric progression of positive real numbers.

Question

Let $A_1, A_2, A_3, \ldots$ be an increasing geometric progression of positive real numbers. If $A_1A_3A_5A_7 = \dfrac{1}{1296}$ and $A_2+A_4 = \dfrac{7}{36}$, then the value of $A_6+A_8+A_{10}$ is equal to:

Accepted Answer

$A_1A_3A_5A_7 = A_4^4 = \dfrac{1}{1296} \Rightarrow A_4 = \dfrac16$ (positive, increasing G.P.).

$A_2+A_4 = \dfrac{7}{36} \Rightarrow A_2 = \dfrac{7}{36}-\dfrac16 = \dfrac{1}{36}$.

Ratio\^2 between $A_2$ and $A_4$: $r^2 = A_4/A_2 = 6$. Then $A_6=A_4r^2=1$, $A_8=A_6r^2=6$, $A_{10}=A_8r^2=36$.

$A_6+A_8+A_{10} = 1+6+36 = 43.$

Solutions · 1