Sequence And Series Question 113
Question: For a sequence $\left\langle a_n\right\rangle, a_1=2$ and $\frac{a_{n+1}}{a n}=\frac{1}{3}$. Then $\sum_ {r=1}^{20} a_r$ is
Options:
A) $ \frac{20}{2}[4+19\times 3] $
B) $ 3( 1-\frac{1}{3^{20}} ) $
C) $ 2(1-3^{20}) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
The sequence is a G.P. with common ratio $ \frac{1}{3} $ . Now from $ \frac{a(1-r^{n})}{1-r},\frac{2,[1-{{(1/3)}^{20}}]}{1-(1/3)} $ = $ 3,[ 1-\frac{1}{3^{20}} ] $ .
BETA
