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If `a_1,a_2,……….a_n` are in H.P, then the expression `a_1 a_2+a_2+a_3+…+a_(n-1)a_n` is equal to (A) `n(a_1-a_n)` (B) `(n-1)(a_1-a_n)` (C) `na_1a_n` (D) `(n-1)a_1a_n`

A

`(n-1)a_1a_n`

B

`n(a_1-a_n)`

C

`(n-1)(a_1-a_n)`

D

`na_1a_n`

Text Solution

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The correct Answer is:
A
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