Question

For a sequence $\left\{a_n\right\}$, $a_1=2$ and $\frac{a_{n+1}}{a_n}=\frac{1}{3}$. Then $\sum _{r=1}^{20}{a}_r$ is

• A. $\frac{20}{2}$ $[4+19\times 3]$
• B. $3(1-\frac{1}{3^{20}})$
• C. $2(1-3^{20})$
• D. None of these

Answer 1

The correct option is B $3(1-\frac{1}{3^{20}})$

According to the problem,

$a_1=2$ and $a_{n+1}=\frac{a_n}{3}$

Using these we have

$a_2=\frac{a_1}{3}=\frac{2}{3},a_3=\frac{a_2}{3}=\frac{2}{3^2}$$,.......a_{20}=\frac{2}{3^{19}}$.

Now,

$\sum _{r=1}^{20}{a}_r=(2+\frac{2}{3}+\frac{2}{3^2}+.....+\frac{2}{3^{19}})=2\frac{(1-\frac{1}{3^{20}})}{(1-\frac{1}{3})}=3(1-\frac{1}{3^{20}})$.