Let a_1, a_2, a_3, be a G.P.

Question

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

Accepted Answer

$\begin{gathered} {a}_{2}\cdot {a}_{3}\cdot {a}_{4}=64 \ \frac{a}{r}\cdot a\cdot ar=64 \ \Rightarrow \, a=4 \ {a}_{1}+{a}_{3}+{a}_{5}=\frac{813}{7} \ =\frac{4}{{r}^{2}}+4+4{r}^{2}=\frac{813}{7} \ {r}^{2}=t \ 28{t}^{2}-785t+28=0 \ (28t-1)(t-28)=0 \ \Rightarrow \, t=28,\frac{1}{28} \end{gathered}$

$\begin{gathered} {r}^{2}=28(\because ) \ {a}_{3}+{a}_{5}+{a}_{7} \ =a+a{r}^{2}+a{r}^{4}={r}^{2}(a+\frac{a}{{r}^{2}}+{r}^{2}) \ =28	imes \frac{813}{7} \ =3252 \end{gathered}$

Solutions · 1