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AP-then-GP Sequence of 4n+1 Terms: Middle Term | JEE

JEE Maths question with a full step-by-step solution.

Question
In a sequence of (4n+1)(4n+1) terms, the first (2n+1)(2n+1) terms are in A.P. with common difference 22, and the last (2n+1)(2n+1) terms are in G.P. with common ratio 0.50.5. If the middle terms of the A.P. and the G.P. are equal, then the middle term of the sequence is
An 2n−12n−1\frac{n\,2^{n-1}}{2^n-1}
Bn 2n+122n−1\frac{n\,2^{n+1}}{2^{2n}-1}
Cn 2nn\,2^n
Dn 2n+12n−1\frac{n\,2^{n+1}}{2^n-1}correct
Solution
Step 1: Let the first term be aa. The A.P. is a,a+2,…,a+4na, a+2, \ldots, a+4n, so its middle (the (n+1)(n+1)th) term is a+2na+2n. The middle of the whole (4n+1)(4n+1)-term sequence is the (2n+1)(2n+1)th term, a+4n.a+4n. Step 2: The last (2n+1)(2n+1) terms form a G.P. starting at a+4na+4n with ratio 12\frac12, so the G.P. middle is (a+4n)(12)n(a+4n)\left(\frac12\right)^n. Equate the two middles:
a+2n=(a+4n)(12)n.a+2n=(a+4n)\left(\tfrac12\right)^n.
 Step 3: Solve for aa and form the required middle a+4na+4n:
a+4n=2n1−(12)n=2n⋅2n2n−1=n 2n+12n−1.a+4n=\frac{2n}{1-\left(\frac12\right)^n}=\frac{2n\cdot 2^n}{2^n-1}=\frac{n\,2^{n+1}}{2^n-1}.
 Correct answer: (4)
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