Question

If the first term of a geometric series is $2$, and its third term is $8$, how many digits are there in the ${40}^{th}$ term of the series, if the common ratio of the sequence is positive?

• A. $10$
• B. $11$
• C. $12$
• D. $13$
• E. $14$

Answer 1

Answer: (D)

$13$

The first number of geometric series is $2$ and third is $8$, let $r$ be common factor

$8=2(r{)}^2$

$\Rightarrow r=2$

$\Rightarrow a_n=a{r}^{n-1}$

$\Rightarrow a_{40}=2\times (2{)}^{40-1}=2\times 2^{39}=13$ digit number