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. $(1-\frac{1}{3^{20}})$

Answer 1

Answer: (B)

$3(1-\frac{1}{3^{20}})$

Given that,

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

${\sum }_{r=1}^{20}{a}_r=\frac{a(1-r^n)}{1-r}$

$=\frac{2[1-{\left(\frac{1}{3}\right)}^{20}]}{1-\frac{1}{3}}$ $[n=20]$

$=\frac{2[1-{\left(\frac{1}{3}\right)}^{20}]}{\frac{2}{3}}$

$=3[1-{\left(\frac{1}{3}\right)}^{20}]$

Hence this is the correct answer.