edited by
1,163 views

1 Answer

Best answer
7 votes
7 votes

It is an infinite G.P. with first term $a=1$ and common ratio $r = \dfrac{1}{4}$.

Sum of infinite G.P with $|r| < 1$ is

$S_{\infty}=\dfrac{a}{1-r}$

$S_{\infty}=\dfrac{1}{1-\dfrac{1}{4}}$

$S_{\infty}=\dfrac{4}{3}$

Hence option d) is correct

edited by
Answer:

Related questions

5 votes
5 votes
2 answers
2
5 votes
5 votes
1 answer
4