If a 1- a 2- a 3- a n are in AP- where a i -0 for all i- then the value of 1- a 1- a 2-1- a 2- a 3-1- a n -2- a n isA- 1-a1-anB- 1-a1-anC- n-a1-anD- n -1-a1- a n

The correct option is D (n 1)/√(a_1)+√(a_n)1/√(a_1)+√(a_2)+1/√(a_2)+√(a_3)+… + 1/√(a_n 1)+√(a_n) =1/√(a_1)+√(a_2)×√(a_1) √(a_2)/√(a_1) √(a_2)+1/√(a_2)+√(a_3)× ...

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Quantitative Aptitude

Progressions

If a 1, a 2, ...

Question

If $a_{1}, a_{2}, a_{3} \dots, a_{n}$ are in AP, where $a_{i} > 0$ for all i, then the value of $\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \dots + \frac{1}{\sqrt{a_{n - 2}} + \sqrt{a_{n}}}$ is

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{n}}}

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\frac{1}{\sqrt{a_{1}} - \sqrt{a_{n}}}

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\frac{n}{\sqrt{a_{1}} + \sqrt{a_{n}}}

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

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Solution

The correct option is D $\frac{(n - 1)}{\sqrt{a_{1}} + \sqrt{a_{n}}}$
$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \dots + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} = \frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} \times \frac{\sqrt{a_{1}} - \sqrt{a_{2}}}{\sqrt{a_{1}} - \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} \times \frac{\sqrt{a_{2}} - \sqrt{a_{3}}}{\sqrt{a_{2}} - \sqrt{a_{3}}} + \dots + \frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} \times \frac{\sqrt{a_{n - 1}} - \sqrt{a_{n}}}{\sqrt{a_{n - 1}} - \sqrt{a_{n}}} = \frac{\sqrt{a_{1}} - \sqrt{a_{2}}}{\sqrt{a_{1}} - \sqrt{a_{2}}} + \frac{\sqrt{a_{2}} - \sqrt{a_{3}}}{\sqrt{a_{2}} - \sqrt{a_{3}}} + \dots + \frac{\sqrt{a_{n - 1}} - \sqrt{a_{n}}}{\sqrt{a_{n - 1}} - \sqrt{a_{n}}} = \frac{\sqrt{a_{1}} - \sqrt{a_{2}}}{- d} + = \frac{\sqrt{a_{2}} - \sqrt{a_{3}}}{- d} + \dots + \frac{\sqrt{a_{n - 1}} - \sqrt{a_{n}}}{- d}$
[d= common difference]
$= \frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{d} + \frac{\sqrt{a_{3}} - \sqrt{a_{2}}}{d} + \dots + \frac{\sqrt{a_{n}} - \sqrt{a_{n - 1}}}{d} = \frac{1}{d} [\sqrt{a_{2}} - \sqrt{a_{1}} + \sqrt{a_{3}} - \sqrt{a_{2}} + \dots + \sqrt{a_{n}} - \sqrt{a_{n - 1}}] = \frac{1}{d} [- \sqrt{a_{1}} + \sqrt{a_{n}}]$ (Rest all terms will cancelled)
$= \frac{1}{d} [- \sqrt{a_{1}} + \sqrt{a_{n}}] = \frac{1}{d} [\sqrt{a_{n}} - \sqrt{a_{1}}] = \frac{1}{d} \sqrt{a_{n}} - \sqrt{a_{1}} \times \frac{\sqrt{a_{n}} + \sqrt{a_{1}}}{\sqrt{a_{n}} + \sqrt{a_{1}}} = \frac{1}{d} \frac{a_{n} - a_{1}}{\sqrt{a_{n}} + \sqrt{a_{1}}} = \frac{1}{d} \frac{a_{1} + (n - 1) d - a_{1}}{\sqrt{a_{n}} + \sqrt{a_{1}}} = \frac{1}{d} \frac{(n - 1) d}{\sqrt{a_{n}} + \sqrt{a_{1}}} = \frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n}}}$

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Similar questions

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

Q. 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}}}}

If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are in arithmetic progression, where $a_{1} > 0$ for all i.
Prove that $\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \dots + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} = \frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n}}}$

Q. If

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}}

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If a1,a2,a3…,an are in AP, where ai>0 for all i, then the value of 1√a1+√a2+1√a2+√a3+…+1√an−2+√an is

If $a_{1}, a_{2}, a_{3} \dots, a_{n}$ are in AP, where $a_{i} > 0$ for all i, then the value of $\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \dots + \frac{1}{\sqrt{a_{n - 2}} + \sqrt{a_{n}}}$ is