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Question: If a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>….a<sub>2n + 1</sub> are in A.P. then \(\frac{a_{2n + ...

If a1, a2, a3….a2n + 1 are in A.P. then a2n+1−a1a2n+1+a1\frac{a_{2n + 1} - a_{1}}{a_{2n + 1} + a_{1}} + a2n−a2a2n+a2\frac{a_{2n} - a_{2}}{a_{2n} + a_{2}}+.....an+2−anan+2+an\frac{a_{n + 2} - a_{n}}{a_{n + 2} + a_{n}}

A

n(n+1)2.a2−a1an+1\frac{n(n + 1)}{2}.\frac{a_{2} - a_{1}}{a_{n + 1}}

B

n(n+1)2\frac{n(n + 1)}{2}

C

(n + 1) (a2 – a1)

D

None of these

Answer

n(n+1)2.a2−a1an+1\frac{n(n + 1)}{2}.\frac{a_{2} - a_{1}}{a_{n + 1}}

Explanation

Solution

The general term can be given by

tr + 1 = a2n+1−r−ar+1a2n+1−r+ar+1\frac{a_{2n + 1 - r} - a_{r + 1}}{a_{2n + 1 - r} + a_{r + 1}}, r = 0, 1, 2, ……, n – 1

= a1+(2n−r)d−{a1+rd}a1+(2n−r)d+{a1+rd}\frac{a_{1} + (2n - r)d - \{ a_{1} + rd\}}{a_{1} + (2n - r)d + \{ a_{1} + rd\}} = (n−r)da1+nd\frac{(n - r)d}{a_{1} + nd}

\ The required sum

dn = ∑r=0n−1tr+1\sum_{r = 0}^{n - 1}t_{r + 1} = ∑r=0n−1(n−r)da1+nd\sum_{r = 0}^{n - 1}\frac{(n - r)d}{a_{1} + nd}= [n+(n−1)+(n−2)+....+1a1+nd]\left\lbrack \frac{n + (n - 1) + (n - 2) + .... + 1}{a_{1} + nd} \right\rbrackd= n(n+1)d2an+1\frac{n(n + 1)d}{2a_{n + 1}} = n(n+1)2\frac{n(n + 1)}{2}.a2−a1an+1\frac{a_{2} - a_{1}}{a_{n + 1}}. [Q d = a2 – a1]