If a_1, a_2, a_3, ..., a_n is an arithmetic p | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the common difference of the given arithmetic progression be $d$.Then,$a_2 - a_1 = a_3 - a_2 = ... = a_n - a_{n-1} = d$.Now,consider the sum $

Vedclass · Accepted Answer

$\frac{n - 1}{a_1 a_n}$

Vedclass · Answer

(D) Let the common difference of the given arithmetic progression be $d$.Then,$a_2 - a_1 = a_3 - a_2 = ... = a_n - a_{n-1} = d$.Now,consider the sum $S = \frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + ... + \frac{1}{a_{n-1} a_n}$.Multiplying and dividing by $d$,we get $S = \frac{1}{d} [\frac{d}{a_1 a_2} + \frac{d}{a_2 a_3} + ... + \frac{d}{a_{n-1} a_n}]$.Substituting $d = a_{k+1} - a_k$,we get $S = \frac{1}{d} [(\frac{1}{a_1} - \frac{1}{a_2}) + (\frac{1}{a_2} - \frac{1}{a_3}) + ... + (\frac{1}{a_{n-1}} - \frac{1}{a_n})]$.This is a telescoping sum,so $S = \frac{1}{d} [\frac{1}{a_1} - \frac{1}{a_n}] = \frac{1}{d} [\frac{a_n - a_1}{a_1 a_n}]$.Since $a_n = a_1 + (n-1)d$,we have $a_n - a_1 = (n-1)d$.Therefore,$S = \frac{1}{d} [\frac{(n-1)d}{a_1 a_n}] = \frac{n-1}{a_1 a_n}$.

If $a_1, a_2, a_3, ..., a_n$ is an arithmetic progression,then $\frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \frac{1}{a_3 a_4} + ... + \frac{1}{a_{n-1} a_n} = ...$

Similar Questions

The sum of the series $(1^2 + 1) \cdot 1! + (2^2 + 1) \cdot 2! + (3^2 + 1) \cdot 3! + \dots + (n^2 + 1) \cdot n!$ is:

The expression $\frac{2^2+1}{2^2-1}+\frac{3^2+1}{3^2-1}+\frac{4^2+1}{4^2-1}+\ldots+\frac{(2011)^2+1}{(2011)^2-1}$ lies in the interval

Find the sum to $n$ terms of the series $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots$

The sum of $10$ terms of the series $\frac{3}{1^{2} \times 2^{2}}+\frac{5}{2^{2} \times 3^{2}}+\frac{7}{3^{2} \times 4^{2}}+\ldots$ is :

The value of $1000 \left[ \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots + \frac{1}{999 \times 1000} \right]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module