Question

The sum of first two terms of a G.P is 5 and each term is 3 times the sum of following terms, then the $4^{th}$ term of the series is:

• A. $4$
• B. $\frac{1}{4}$
• C. $\frac{1}{16}$
• D. $\frac{1}{32}$

Answer 1

The correct option is A $4$

$a_n=3(9_{n-1}+9_{n+2}+........)$

$a{r}^{n-1}=3(a{r}^n+a{r}^{n+1}+a{r}^{n+2}+......)$

$a{r}^{n-1}=\frac{3a{r}^n}{1-r}$

$\frac{3r}{1-r}=1$

$3r=1-r$

$4r=1$

$r=\frac{1}{4}$

When $r=\frac{1}{4}$, we get

$a+\frac{a}{4}=5$

$\frac{5a}{4}=5$

$G.P=4,1,\frac{1}{4},\frac{1}{16}$

$a=4$