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

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Proof by mathematical induction

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

Since for all i, $a_{i} > 0$ , therefore above result can be proved by using principle of mathematical induction.
Step 1: Checking for n=2,
$\frac{1}{a_{1} a_{2}} = \frac{1}{a_{1} a_{2}}$ which is true.
Step 2: Assuming the given relation to be true for $1 <= m < n$ ,
$\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} +$ so on upto $+ \frac{1}{a_{m - 1} a_{m}} = \frac{m - 1}{a_{1} a_{m}}$

Step 3: To prove that result is true for next natural number m+1.
let d be the common difference
we have $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} +$ so on upto $\frac{1}{a_{m - 1} a_{m}} + \frac{1}{a_{m} a_{m + 1}} = \frac{m - 1}{a_{1} a_{m}} + \frac{1}{a_{m} a_{m + 1}}$

$\to \frac{1}{a_{m}} * \frac{m a_{m + 1} - a_{m + 1} - a_{1}}{a_{1} a_{m + 1}}$ $\to \frac{1}{a_{m}} \frac{m a_{m + 1} - (a_{1} + m d) - a_{1}}{a_{1} a_{m + 1}}$

on further simplification, we get $\frac{m}{a_{1} a_{m + 1}}$ which is true for m+1 and hence for all n.
Hence $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} +$ so on upto $+ \frac{1}{a_{n - 1} a_{n}} = \frac{n - 1}{a_{1} a_{n}}$

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