Question

Find the sum of the infinitely decreasing G.P whose third term, the triple product of the first term by the fourth term and the second term form in the indicated order, an A.P with the common difference equal to 1/8.

Answer 1

Let G.P be $a,ar,a{r}^2,...$ since the G.P is infinite and decreasing
$-1<r<1$ and $r>0,$ so
$a>0$. According to the hypothesis,
$a{r}^2,3a.a{r}^3,ar$ are in A.P with common difference $\frac{1}{8}$
$\Rightarrow 3a^2{r}^3-a{r}^2=\frac{1}{8}\cdots \left(1\right)$
and $ar=a{r}^2+2.\frac{1}{8}\cdots \left(2\right)$
from $\left(2\right)$,
$a=\frac{1}{4(r-r^2)}$
$a=\frac{1}{4r(1-r)}\cdots \left(3\right)$
from $\left(1\right)$ and $\left(3\right)$
$\frac{3r^3}{16r^2(1-r{)}^2}-\frac{r^2}{4r(1-r)}=\frac{1}{8}$
$\Rightarrow \frac{3r}{16(1-r{)}^2}-\frac{r}{4(1-r)}=\frac{1}{8}$
$\therefore 2r^2+3r-2=0$
which gives, $r=\frac{1}{2},-2$, but $\therefore r=\frac{1}{2}$
from $\left(3\right)$, $a=1$
Hence G.P is $1,\frac{1}{2},\frac{1}{4},\frac{1}{8},.....$
$\therefore S_{\infty }=\frac{1}{1-\frac{1}{2}}$
$=2$