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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_(r=1)^(k)(1)/(x_(i))` is equal to

A

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

B

`(75)/(k)`

C

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

D

`(75)/(n)`

Text Solution

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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