In a sequence of `(4n+1)` terms, the first `(2n+1)` terms are n 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 LG.P. are equal ,then the middle terms of the sequence is `(n .2 n+1)/(2^(2n)-1)` b. `(n .2 n+1)/(2^n-1)` c. `n .2^n` d. none of these

Options

A. `(n.2^(n+1))/(2^n-1)`
B. `(n.2^(2+1))/(2^(2n-1))`
C. `n2^n`
D. none of these

Answer

A

Explanation

`a_(n+1) = a_(2n+1) - nd =a_(2n+1) - 2n" "a_(3n+1) =a_(2n+1)(r)^n = a_(2n+1)(1/2)^n`
Since `a_(n +1) = a_(3n+1)`
`a_(2n +1) -2n = a_(2n +1)(1/2)^n :.a_(2n +1) = (n. 2^(n +1))/(2^n -1)`