Question

If the sum to infinity of the series $3+(3+d)\frac{1}{4}+(3+2d)\frac{1}{4^2}+\cdot \cdot \cdot$ is $\frac{44}{9}$, then find $d$.

• A. $1$
• B. $2$
• C. $-1$
• D. $4$

Answer 1

Answer: (B)

$2$

$\frac{44}{9}=3+(3+d)\frac{1}{4}+(3+2d)\frac{1}{4^2}+\dots$......... {i}

By dividing above equation by $4$,

$\frac{11}{9}=0+\left(3\right)\frac{1}{4}+(3+d)\frac{1}{4^2}+(3+2d)\frac{1}{4^3}+\dots$............ {ii}

Subtracting {ii} from {i} term by term, we get

$\frac{33}{9}=3+\frac{d}{4}+\frac{d}{4^2}+\frac{d}{4^3}+\dots$

Using the formula for infinite G.P. $(|r|<1)$

$a+ar+a{r}^2+\cdots =\frac{a}{1-r}$

We get,

$\frac{33}{9}=3+\frac{\frac{d}{4}}{1-\frac{1}{4}}$

$\Rightarrow \frac{33}{9}-3=\frac{d}{3}$

$\Rightarrow \frac{2}{3}\times 3=d$

$\Rightarrow d=2$