If \( a_{1}, a_{2}, a_{3}, \ldots a_{2 n+1} \), are in A.P., then \( \frac{a_{2 n+1}-a_{1}}{a_{2...
If \( a_{1}, a_{2}, a_{3}, \ldots a_{2 n+1} \), are in A.P., then \( \frac{a_{2 n+1}-a_{1}}{a_{2 n+1}+a_{1}}+\frac{a_{2 n}-a_{2}}{a_{2 n}+a_{2}}+\ldots . .+\frac{a_{n+2}-a_{n}}{a_{n+2}+a_{n}} \) is equal to
(a) \( \frac{n(n+1)}{2} \frac{\left(a_{2}-a_{1}\right)}{a_{n+1}} \)
(b) \( \frac{n(n+1)}{2} \)
(c) \( \frac{(n-1) n\left(a_{2}-a_{1}\right)}{a_{n+1}} \)
(d) \( \frac{n(n-1)(n+1)\left(a_{2}-a_{1}\right)}{a_{n+1}} \)
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