Question

In a sequence of $21$ terms, the first $11$ term 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 the $A.P.$ is equal to the middle term of $G.P.$ The find the middle term of the entire sequence.

Answer 1

The first 11 term are in A.P with the common difference $d=2$ and the first term is $a$

Then the middle term of A.P is the $6^{th}$ term i.e,

$a_6=a+(6-1)d=a+5.2=a+10$

In case of G.P, first term is $p$ and the common ratio $r=2$.

Then the middle term of G.P is the $6^{th}$ term of the G.P can be written as

$p_6=p{r}^{6-1}=p{2}^5$

Acc. to Qn,

$a_6=p_6\Rightarrow (a+10)=p.2^5⟶\left(1\right)$

also ${11}^{th}$ term of A.P $=$ $1^{st}$ term of G.P

so,

$a_{11}=p_1\Rightarrow a+(11-1).2=p.r^{1-1}\Rightarrow a+20=p⟶\left(2\right)$

solving (1) and (2)

$p=\frac{-10}{31}anda=\frac{-630}{31}$

therefore middle term of the sequence is ${11}^{th}$ of A.P or $1^{st}$ term of G.P i.e, $p=\frac{-10}{31}$