Question

In a sequence of (4n+1) terms, the first (2n+1) terms are in A.P., whose common difference is 2, and the last (2n+1) terms are in G.P whose common ratio is 0.5 if the middle terms of the A.P and G.P are equal then the middle term of the sequence is

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Let the first term is a , then first (2n+1) terms are
a, a + 2, a + 4,.....a + 2.2n. Clearly the middle term of the sequence of 4n + 1 term is $(2n+1{)}^{th}$ term, i.e . a + 4n also the middle term of the A.P of (2n + 1) term is $(n+1{)}^{th}$ term i.e., a + 2n. Again for the last (2n + 1) terms the first term will be $(2n+1{)}^{th}$ term of the A.P i.e. a + 4n
$\therefore GPis(a+4n),(1+4n)(0.5{)}^n$
Its middle term is $(a+4n)(0.5{)}^n$
According to the given condition,
$a+2n=(1+4n)(0.5{)}^n$
$\therefore a=\frac{2n-4n(0.5{)}^n}{(0.5{)}^n-1}$
Required middle term =a + 4n =
$\frac{2n-4n(0.5{)}^n}{(0.5{)}^n-1}+4n=\frac{2n}{1-{\left(\frac{1}{2}\right)}^2}=\frac{n{.2}^{n+1}}{2^n-1}$