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 terms the A.P. is equal to the middle term of the G.P., then the middle term of the entire sequence is

• A. $-10/31$
• B. $10/31$
• C. $32/31$
• D. $-31/32$

Answer 1

The correct option is A $-10/31$

The middle term of $AP=6^{th}$ term

$a+(n-1)d$

$=a+10$

The middle term of $GP={16}^{th}$

$=P{r}^{n-1}$

$=P{2}^{6-1}=P{2}^5$

$a+10=P{.2}^5$ -------- $\left(1\right)$

${11}^{th}$ term of $ap=1^{st}$ term of $GP$

$a+10.2=P$

$a+20=P$------ $\left(2\right)$

$P=-10/31$ $a=-630/31$

middle term of sequence $={11}^{th}$ term $=\frac{-10}{31}$