In a sequence of $(4 n + 1)$ terms the $1^{st}$ $(2 n + 1)$ terms are in A.P. whose common difference is 2 and the last $(2 n + 1)$ terms are in G.P. whose common ratio is $\frac{1}{2}$ . If the middle terms of the A.P. and G.P. are equal, then Middle term of the sequence is

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$n 2^{n}$

$\frac{n {.2}^{n + 1}}{2^{n} - 1}$

$\frac{n {.2}^{n + 1}}{2^{2 n} - 1}$

$(n + 1) 2^{n + 1}$

answer is A.

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Detailed Solution

1^st $(2 n + 1)$ terms of A.P. are $A$ , $A + 2$ , …., $A + 4 n$ .

Last $(2 n + 1)$ terms of G.P. are $(A + 4 n)$ , $(A + 4 n) \frac{1}{2}$ , ……., $(A + 4 n) \frac{1}{2^{2 n}}$

= $A + 2 n = \frac{A + 4 n}{2^{n}}$ Þ $A = \frac{4 n - 2 n 2^{n}}{2^{n} - 1}$

Middle term of sequence = $T_{2 n + 1} = A + 4 n = \frac{n 2^{n + 1}}{2^{n} - 1}$

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