Question

In a sequence of 4n+1 terms, the first 2n+1 terms are in a.p. whose d is 2 and the last 2n+1 terms are in g.p. whose r is 0.5 , if the middle terms of a.p. and g.p. are equal, then middle term of sequence in terms of only n.

Answer 1

Let the first term be a
then a, a+2, a+4, ...a+2,2n are in A.P.
middle term of sequence is $(2n+1{)}^{th}$ term i.e., a+4n
(2n+1) terms in AP
$\therefore$ Middle term of AP $=(n+1{)}^{th}$ term
= a+2n
for last (2n+1) terms the first term will be $(2n+1{)}^{th}$ term of AP i.e., a+4n
$\therefore$ GP is $(a+4n),(1+4n),(0.5{)}^n$
Its middle term is $a+4n\left)\right(0.5{)}^n$
$a=\frac{2n-4n(0.5{)}^n}{(0.5{)}^n-1}$
Required middle term = a+4n
$=\frac{n{.2}^n-1}{2^n-1}$.