If a_r 0, r in N and a_1, a_2, a_3, ..., a_ | vedclass

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

Question

(B) In an arithmetic progression,the sum of terms equidistant from the beginning and end is constant.$a_1 + a_{2n} = a_2 + a_{2n-1} = ... = a_n + a_{n

Vedclass · Accepted Answer

$\frac{n(a_1 + a_{2n})}{\sqrt{a_1} + \sqrt{a_{n+1}}}$

Vedclass · Answer

(B) In an arithmetic progression,the sum of terms equidistant from the beginning and end is constant.$a_1 + a_{2n} = a_2 + a_{2n-1} = ... = a_n + a_{n+1} = k$Each term in the series is of the form $\frac{k}{\sqrt{a_i} + \sqrt{a_{i+1}}}$.Rationalizing the denominator: $\frac{k(\sqrt{a_i} - \sqrt{a_{i+1}})}{a_i - a_{i+1}} = \frac{k(\sqrt{a_i} - \sqrt{a_{i+1}})}{-d}$,where $d$ is the common difference.The sum becomes $\sum_{i=1}^{n} \frac{k(\sqrt{a_i} - \sqrt{a_{i+1}})}{-d} = \frac{k}{-d} (\sqrt{a_1} - \sqrt{a_{n+1}})$.Since $a_{n+1} = a_1 + nd$,we have $a_1 - a_{n+1} = -nd$.Substituting $k = a_1 + a_{2n}$ and the difference,the expression simplifies to $\frac{n(a_1 + a_{2n})}{\sqrt{a_1} + \sqrt{a_{n+1}}}$.

If $a_r > 0, r \in N$ and $a_1, a_2, a_3, ..., a_{2n}$ are in an arithmetic progression,then $\frac{a_1 + a_{2n}}{\sqrt{a_1} + \sqrt{a_2}} + \frac{a_2 + a_{2n-1}}{\sqrt{a_2} + \sqrt{a_3}} + \frac{a_3 + a_{2n-2}}{\sqrt{a_3} + \sqrt{a_4}} + ... + \frac{a_n + a_{n+1}}{\sqrt{a_n} + \sqrt{a_{n+1}}} = ?$

Similar Questions

Let $T_r$ be the $r$-th term of an arithmetic progression for $r = 1, 2, 3, \dots$. If for some positive integers $m$ and $n$,$T_m = \frac{1}{n}$ and $T_n = \frac{1}{m}$,then $T_{mn} = \dots$

If $a, b, c, d, e, f$ are in an arithmetic progression,then $e - c = \dots$

Suppose that all the terms of an arithmetic progression $(A.P.)$ are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6:11$ and the seventh term lies between $130$ and $140$,then the common difference of this $A.P.$ is

If the roots of the equation $x^3 - 9x^2 + \alpha x - 15 = 0$ are in $A.P.$,then $\alpha$ is:

Let $S_{1}$ be the sum of the first $2n$ terms of an arithmetic progression. Let $S_{2}$ be the sum of the first $4n$ terms of the same arithmetic progression. If $(S_{2} - S_{1})$ is $1000$,then the sum of the first $6n$ terms of the arithmetic progression is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module