If a,b,c are in A.P., then frac{1}{{sqrt a + | vedclass

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

Question

(a) $a, b, c$ are in $A.P.$   $i.e.$, $2b = a + c$

Let $\frac{1}{{\sqrt a + \sqrt c }} - \frac{1}{{\sqrt a + \sqrt b }} = \frac{1}{{\sqrt

Vedclass · Accepted Answer

$A.P.$

Vedclass · Answer

(a) $a, b, c$ are in $A.P.$   $i.e.$, $2b = a + c$

Let $\frac{1}{{\sqrt a + \sqrt c }} - \frac{1}{{\sqrt a + \sqrt b }} = \frac{1}{{\sqrt b + \sqrt c }} - \frac{1}{{\sqrt a + \sqrt c }}$

==> $\frac{{\sqrt b - \sqrt c }}{{\sqrt a + \sqrt b }} = \frac{{\sqrt a - \sqrt b }}{{\sqrt b + \sqrt c }}$

$ \Rightarrow b - c = a - b$

==>$2b = a + c$

$	herefore \frac{1}{{\sqrt a + \sqrt b }},\frac{1}{{\sqrt a + \sqrt c }},\frac{1}{{\sqrt b + \sqrt c }}$ are in $A.P.$

If $a,b,c$ are in $A.P.$, then $\frac{1}{{\sqrt a + \sqrt b }},\,\frac{1}{{\sqrt a + \sqrt c }},$ $\frac{1}{{\sqrt b + \sqrt c }}$ are in

Similar Questions

Let $S_n$ denote the sum of first $n$ terms an arithmetic progression. If $S_{20}=790$ and $S_{10}=145$, then $S_{15}-$ $S_5$ is:

If three positive numbers $a, b$ and $c$ are in $A.P.$ such that $abc\, = 8$, then the minimum possible value of $b$ is

If ${\log _5}2,\,{\log _5}({2^x} - 3)$ and ${\log _5}(\frac{{17}}{2} + {2^{x - 1}})$ are in $A.P.$ then the value of $x$ is :-

The sum of the first four terms of an $A.P.$ is $56 .$ The sum of the last four terms is $112.$ If its first term is $11,$ then find the number of terms.

The sum of $24$ terms of the following series $\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} + .........$ is