Let a1- a2- a3 - an be an AP- then- 1-a1 an - 1-a2 aa - 1 - 1-a3 an - 2 - - - 1-an a1 -

Since a_1,a_2,a_3,.......a_n are in AP a_1 + a_n = a_2 + a_n 1 = a_3 + a_n 2..... Consider LHS =1/a_1 + a_n [ a_1 + a_n/a_1a_n + a_2 + a_n/a_2 ...

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Sum of n Terms

Let a1, a2,...

Question

Let $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ be an AP, then.
$\frac{1}{a_{1} a_{n}} + \frac{1}{a_{2} a_{a - 1}} + \frac{1}{a_{3} a_{n - 2}} + . . . . . + \frac{1}{a_{n} a_{1}} =$

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

\frac{1}{a_{1} a_{n}} + \frac{1}{a_{2} a_{n - 1}} + \frac{1}{a_{3} a_{n - 2}} + . . . . . \frac{1}{a_{n} a_{1}} = \frac{2}{(a_{1} + a_{n})} [\frac{1}{a_{1}} + \frac{1}{a_{2}} + . . . . \frac{1}{a_{n}}]

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

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

a_{1} + a_{n} = a_{r} + a_{n - r}

1 \leq r \leq n

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

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

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

+ \frac{1}{a_{n} a_{1}} = \frac{2}{a_{1} + a_{n}} (\frac{1}{a_{1}} + \frac{1}{a_{2}} + . . . + \frac{1}{a_{n}})

If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ be an AP of nonzero terms then prove that

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

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