If a1- a2- a3- -an are in A-P- and ai - 0 for all i- then show that 1-a1a2 - 1-a2a3 - - - 1-an - 1an - n - 1-a1an-

a_1,a_2,....a_n are in A.PLet d be the common difference. 1a_2a_2+1a_2a_3+..+1a_n 1a_n =1d(1a_1 1a_2+1a_2 1a_3_......1a_n 1 1a_n) =1d( ...

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Byju's Answer

Standard XII

Mathematics

Harmonic Progression

If a1, a2, ...

Question

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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Solution

$a_{1}, a_{2}, . . . . a_{n}$ are in A.P
Let $d$ be the common difference.
$\frac{1}{a_{2} a_{2}} + \frac{1}{a_{2} a_{3}} + . . + \frac{1}{a_{n - 1} a_{n}}$

$= \frac{1}{d} (\frac{1}{a_{1}} - \frac{1}{a_{2}} + \frac{1}{a_{2}} - {\frac{1}{a_{3}}}_{.} . . . . . \frac{1}{a_{n - 1}} - \frac{1}{a_{n}})$

$= \frac{1}{d} (\frac{1}{a_{1}} - \frac{1}{a_{n}})$

$= \frac{1}{d} (\frac{1}{a_{1}} - \frac{1}{a_{n}})$

$= \frac{1}{d} (\frac{a_{n} - a_{1}}{a_{1} a_{n}})$

$= \frac{1}{d} (\frac{a_{1} + (n - 1) d - a_{1}}{a_{1} a_{n}})$ $| ∵ a_{n} = a_{1} + (n - 1) d$

$= \frac{1}{d} \frac{(n - 1) d}{a_{1} a_{n}}$

$= \frac{n - 1}{a_{1} a_{n}}$

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are in A.P. and

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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 $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ are in A.P. Where $a_{i} > 0$ for all i, then the value of
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If a1,a2,a3,...,an are in A.P. and ai>0 for all i, then show that 1a1a2+1a2a3+...+1an−1an=n−1a1an.

a1,a2,....an are in A.PLet d be the common difference.1a2a2+1a2a3+..+1an−1an=1d(1a1−1a2+1a2−1a3......1an−1−1an)=1d(1a1−1an)=1d(1a1−1an)=1d(an−a1a1an)=1d(a1+(n−1)d−a1a1an) |∵an=a1+(n−1)d=1d(n−1)da1an=n−1a1an

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