Question

In a sequence of $21$ terms ,the first are in A.P. with common different $2$ and the last $11$ terms are in G.P with common ratio $2$.If the middle terms of the A.P is equal to the middle term of G.P.Then find the middle term of the entire sequence ?

Answer 1

The first $11$ term are in $AP$ with common difference $d=2$ and first term is $a$

then the middle term of $AP$ is the $6^{th}$ term i.e,

$a+(n-1)d=a+(6-1)2=a+10$

Let the $GP$ first term is $P$ and common ratio $r=2$

Then the middle term og ##GP$

i.e, ${16}^{th}$ term of sequence or $6^{th}$ term of $GP$ can be written as $P{r}^{n-1}=P{2}^{6-1}=p{2}^5$

A/Q, $a+10=P{2}^5⟶\left(1\right)$

also, $11$ term of $AP$ $=$ $1$ term of $GP$

$\Rightarrow a+(11-1)2=P\Rightarrow a+20=P⟶\left(2\right)$

solving (1) and (2) we get

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

Middle trem of the sequence is the $11$ term of $AP$ or the $1$ term of $GP$ i.e $P=\frac{-10}{31}$