# if a,b,c,d are an arithmetic progression (in that order), prove that (1)/(sqrt a+sqrt b)+(1)/(sqrt b +sqrt c) +(1)/(sqrt c + \sqrt d) = (3)/(sqrt a + sqrt d)

proving that $\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{c}+\sqrt{d}}=\frac{3}{\sqrt{a}+\sqrt{d}}$ for any A.P.
if $a,b,c,d$ are an arithmetic progression (in that order), prove that
I made $n$ the common difference of $a,b,c,d$; so
$a=a$
$b=a+n$
$c=a+2n$
$d=a+3n$
I tried to replace the terms with those, anyways i squared both equalities but i didn 't get nothing since i'm pretty bad with square roots. I'm looking for some hints or properties that can be useful. Thanks
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Kaiden Weeks
HINT
Multiply and divide by conjugate of each denominator, then you'll get a $\left(-n\right)$ in each denominator.
Then :

$=\frac{\sqrt{a}-\sqrt{d}}{-n}=\frac{a-d}{-n\cdot \left(\sqrt{a}+\sqrt{d}\right)}=\frac{3}{\sqrt{a}+\sqrt{d}}$

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Raynor2i
$\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{c}+\sqrt{d}}=\frac{3}{\sqrt{a}+\sqrt{d}}$
Make into A.P.
$S=\frac{1}{\sqrt{a}+\sqrt{a+x}}+\frac{1}{\sqrt{a+x}+\sqrt{a+2x}}+\frac{1}{\sqrt{a+2x}+\sqrt{a+3x}}-\frac{3}{\sqrt{a}+\sqrt{a+3x}}$
Rationalize
$\frac{1}{\sqrt{a}+\sqrt{a+x}}\frac{\sqrt{a}-\sqrt{a+x}}{\sqrt{a}-\sqrt{a+x}}=\frac{\sqrt{a}-\sqrt{a+x}}{-x}$
$\frac{1}{\sqrt{a+x}+\sqrt{a+2x}}\frac{\sqrt{a+x}-\sqrt{a+2x}}{\sqrt{a+x}-\sqrt{a+2x}}=\frac{\sqrt{a+x}-\sqrt{a+2x}}{-x}$
$\frac{1}{\sqrt{a+2x}+\sqrt{a+3x}}\frac{\sqrt{a+2x}-\sqrt{a+3x}}{\sqrt{a+2x}-\sqrt{a+3x}}=\frac{\sqrt{a+2x}-\sqrt{a+3x}}{-x}$
$\frac{3}{\sqrt{a}+\sqrt{a+3x}}\frac{\sqrt{a}-\sqrt{a+3x}}{\sqrt{a}-\sqrt{a+3x}}=\frac{3\sqrt{a}-3\sqrt{a+3x}}{-3x}=\frac{\sqrt{a}-\sqrt{a+3x}}{-x}$
Combine
$\begin{array}{}S& =\left(-1/x\right)\left(\left(\sqrt{a}-\sqrt{a+x}\right)+\left(\sqrt{a+x}-\sqrt{a+2x}\right)\\ & \phantom{\rule{2em}{0ex}}+\left(\sqrt{a+2x}-\sqrt{a+3x}\right)-\left(\sqrt{a}-\sqrt{a+3x}\right)\right)\\ & =0\phantom{\rule{2em}{0ex}}\text{because everything cancels out!!!}\end{array}$

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