There are (4n+1) terms in a certain sequence of which the first (2n+1) terms form an A.P of common difference 2 and the last (2n +1) terms are in G.P. of common ratio `1//2`. If the middle term of both A.P and G.P. are the same , then find the mid-term of this sequence.

Options

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

Answer

A

Explanation

Let `a_(1),a_(2),a_(3), . . . ,a_(2n),a_(2n+1), . . . ,a_(4n),a_(4n+1)` be the given sequence. Then,
`a_(2n+1)=a_(1)+4nanda_(4n+a)=a_(2n+1)xx((1)/(2))^(2n)`
`rArr" "a_(2n+1)=a_(1)+4nanda_(4n+1)=((a_(1)+2nd))/(2^(2n))`
Also,
Middle term of first (2n+1) terms = Middle term of the last (2n+1) terms
`rArr" "a_(n+1)=a_(3n+1)`
`rArr" "a_(1)+2n=a_(2n+1)((1)/(2))^(n)`
`rArr" "a_(1)+2n=(a_(1)+4n)xx(1)/(2^(n))`
`rArr" "a_(1){1-(1)/(2^(n))}=(4n)/(2^(n))-2n`
`rArr" "a_(1)(2^(n)-1)=4n-2nxx2^(n)rArra_(1)=(4n-2nxx2^(n))/(2^(n)-1)`
`:.` Middle term of the sequence `=a_(2n+1)=a_(1)+4n`
`=(4n-2nxx2^(n))/(2^(n)-1)+4n`
`=(2nxx2^(n))/(2^(n)-1)=(n*2^(n+1))/(2^(n)-1)`