An AP of non-zero terms- Prove that If a 1- a 2- a 3- - a n be1- a 1 a 2-1- a 2 a 3-1- a n -1 a n -n -1- a 1 a n-

Let d be the common difference of the given AP, then a_2 a_1 = a_3 a_2 = ...=a_n a_n 1=d LHS = 1/a_1a_2+1/a_2a_3+1/a_3a_4+...+1/a_n 1 a_n = 1/d ( d/a_1a_2+d/a_ ...

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Standard XI

Mathematics

Summation by Sigma Method

an AP of non-...

Question

If $a_{1}, a_{2}, a_{3}, . . . ., a_{n}$ be an AP of non-zero terms. Prove 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

Let d be the common difference of the given AP, then

$a_{2} - a_{1} = a_{3} - a_{2} = . . . = a_{n} - a_{n - 1} = d$

LHS = $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + . . . + \frac{1}{a_{n - 1} a_{n}}$

$= \frac{1}{d} (\frac{d}{a_{1} a_{2}} + \frac{d}{a_{2} a_{3}} + \frac{d}{a_{3} a_{4}} + . . . + \frac{d}{a_{n - 1} a_{n}})$

$= \frac{1}{d} (\frac{a_{2} - a_{1}}{a_{1} a_{2}} + \frac{a_{3} - a_{2}}{a_{2} a_{3}} + \frac{a_{4} - a_{3}}{a_{3} a_{4}} + . . . . + \frac{a_{n} - a_{n - 1}}{a_{n - 1} - 1_{n}^{a}})$

= $\frac{1}{d} (\frac{1}{a_{1}} - \frac{1}{a_{2}} + \frac{1}{a_{3}} - \frac{1}{a_{3}} - \frac{1}{a_{4}} + . . . + \frac{1}{a_{n - 1} - \frac{1}{a_{n}}})$

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

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

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If a1,a2,a3,....,an be an AP of non-zero terms. Prove that 1a1a2+1a2a3+....+1an−1an= n−1a1an

If $a_{1}, a_{2}, a_{3}, . . . ., a_{n}$ be an AP of non-zero terms. Prove that

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

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