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^(n+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)" " therefore a_(2n+1)=(n*2^(n+1))/(2^(n)-1)`