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

Explanation

we have,d=2, r=`1//2`
There are 4n+1 terms. Then the mid term is (2n+1)th term. `T_(n+1) and t_(n+1)` are mid terms of A.P and G.P.
`T_(n+1)=a+nd=a+2n`
`t_(n+1)=AR^(n)=T(2n+1)xx(1/2)^(n)=(a+4n)(1/2)^(n)`
By given condition,
`T_(n+1=t_(n+1)`
`rArr a+2n=(a+4n)1/(2^(n))`
or `(2^(n)-1)a=4n-2nxx2^(n)`
or `a=(4n-nxx2^(n+1))/(2^(n)-1)`
Hence, the mid-term of the sequence is
`a+4n=(4n-nxx2^(n+1))/(2^(n)-1)`
Hence, the mid-term of the sequence is
`a+4n=(4n-nxx2^(n+1))/(2^(n)-1)+4n`
`=(-nxx2^(n+1)+2nxx2^(n+1))/(2^(n)-1)`
`=(nxx2^(n+1))/(2^(n)-1)`