Question

In a sequence of 21 terms, the first 11 terms are in A.P. with common difference 2 and the last 11 terms are in G.P. with common ratio 2. If the middle term of A.P. be equal to the middle term of the G.P., then the middle term of the entire sequence is

• A. $-\frac{10}{31}$
• B. $\frac{10}{31}$
• C. $\frac{32}{31}$
• D. $-\frac{31}{32}$

Answer 1

Answer: (A)

For the first $11$ terms in $A.P., d = 2$
Middle term of the $A.P.$ is $6$th term,
$ a_{b} = a+5d = a= 10$,
$ a_{11} = a+10d = a+20$
For next $11$ terms in $G.P., r = 2$
Middle term of the $G.P.$ is $6$th terms $ = b\left(2\right)^{5} $
where $b=$ last term of $A.P.$
$ \Rightarrow \left(a+20\right)32 = a+10$
$\Rightarrow 32a+20\times32= a+10 $
$\Rightarrow 31a=10-20\times32$
$\Rightarrow a= -\frac{630}{31} $
$\therefore $ middle term of entire sequence is $11^{th }$ term
$=\frac{-630}{31} +10\times d $
$= \frac{-630}{31} +10 \times2$
$= -\frac{10}{31}$
$ \therefore $ required term is $ \frac{-10}{31}$