Question

In a sequence of 21 tems, 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 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

Middle term of A.P. = Middle term of G.P.
$\Rightarrow a+10=(a+20)2^5$
$\Rightarrow -31a=630$
$\Rightarrow a=-630/31$
Now, middle term of entire sequence is

$a+20=-\frac{630}{31}+20=-\frac{10}{31}$