Question

If $6^{th}$ term of a G.P. is $\frac{1}{32}$ and $9^{th}$ term is $\frac{1}{256}$, then ${11}^{th}$ term $=$

• A. $1024$
• B. $\frac{1}{1024}$
• C. $\frac{1}{256}$
• D. $\frac{1}{512}$

Answer 1

Answer: (B)

$\frac{1}{1024}$

let the first term be a and common ratio be r
then sixth term is $a_6=a.r^5=\frac{1}{32}$
9th term is $a_9=a.r^8=\frac{1}{256}$
now from $\frac{a_9}{a_6}=r^3=\frac{1}{8}$
there fore r = $\frac{1}{2}$
and we also get a= 1 by substituting it in $a_6=\frac{1}{32}$
now $a_{11}=a.r^{10}=\frac{1}{1024}$