Let x(1),x(2),x(3), . . .,x(k) be the di...

Let `x_(1),x_(2),x_(3), . . .,x_(k)` be the divisors of positive integer '`n`' (including `1` and `x`). If `x_(1)+x_(2)+ . . .+x_(k)=75`, then `sum_(i=1)^(k)(1)/(x_(i))` is equal to:

A

(a) `(k^(2))/(75)`

B

(b) `(75)/(k)`

C

(c) `(n^(2))/(75)`

D

(d) `(75)/(n)`

Verified by Experts

The correct Answer is:

D

`underset(i=1)overset(k)(sum)(1)/(x_(i))=(1)/(x_(1))+(1)/(x_(2))+ . . .+(1)/(x_(k))=(underset(i=1)overset(k)(sum)x_(i))/(n)=(75)/(n)`

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