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Substitution Method to Remove Indeterminate Form

If a 1 , a ...

Question

If $a_{1}, a_{2}, a_{3}, . . . ., a_{n}$ are in A.P. where $a_{i} > 0$ for all i, show that $\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . + \frac{1}{\sqrt{a_{n - 1} + \sqrt{a_{n}}}} = \frac{(n - 1)}{\sqrt{a_{1} + \sqrt{a_{n}}}}$

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Solution

L.H.S
$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . .$
$= \frac{1 (\sqrt{a_{1}} - \sqrt{a_{2}})}{(\sqrt{a_{1}} + \sqrt{a_{2}}) (\sqrt{a_{1}} - \sqrt{a_{2}})} + \frac{1 (\sqrt{a_{2}} - \sqrt{a_{3}})}{(\sqrt{a_{2}} + \sqrt{a_{3}}) (\sqrt{a_{2}} - \sqrt{a_{3}})} + . . . . . .$
$= \frac{\sqrt{a_{1}} - \sqrt{a_{2}}}{a_{1} - a_{2}} + \frac{\sqrt{a_{2}} - \sqrt{a_{3}}}{a_{2} - a_{3}} + . . . . . .$
$= \frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{d} + \frac{\sqrt{a_{3}} - \sqrt{a_{2}}}{d} + . . . . . .$
$= \frac{\sqrt{a_{n}} - \sqrt{a_{1}}}{d} \times (\frac{\sqrt{a_{n}} + \sqrt{a_{1}}}{\sqrt{a_{n}} + \sqrt{a_{1}}})$
$= \frac{a_{n} - a_{1}}{(\sqrt{a_{1}} + \sqrt{a_{n}}) d} = \frac{(n - 1)}{\sqrt{a_{1}} + \sqrt{a_{n}}}$

R.H.S
Hence, proved.

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Similar questions

a_{1}, a_{2}, a_{3}, . . ., a_{n}

are in A.P. and

a_{i} > 0

for all i, then show that

\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + . . . + \frac{1}{a_{n - 1} a_{n}} = \frac{n - 1}{a_{1} a_{n}}

a_{1}, a_{2}, a_{3}, . . ., a_{n}

are in A.P. and

a_{i} > 0

for all i, then show that

\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + . . . + \frac{1}{a_{n - 1} a_{n}} = \frac{n - 1}{a_{1} a_{n}}

a_{1}, a_{2},

a_{n}

are in A.P. where

a_{i} > 0

for all ii, show that

\frac{1}{\sqrt{(a_{1})} + \sqrt{(a_{2})}} + \frac{1}{\sqrt{(a_{2})} + \sqrt{(a_{3})}}

+ \frac{1}{\sqrt{(a_{n - 1})} + \sqrt{(a_{n})}} = \frac{n - 1}{\sqrt{(a_{1})} + \sqrt{(a_{n})}}

If $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ are in A.P. Where $a_{i} > 0$ for all i, then the value of
$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} =$

If $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ are in A.P. Where $a_{i} > 0$ for all i, then the value of
$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} =$

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