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Let a(1), a(2),…. and b(1),b(2),…. be ar...

Let `a_(1), a_(2),….` and `b_(1),b_(2),….` be arithemetic progression such that `a_(1)=25`, `b_(1)=75` and `a_(100)+b_(100)=100`, then the sum of first hundred term of the progression`a_(1)+b_(1)`, `a_(2)+b_(2)`,…. is equal to

A

`(n-1)D_(n-1)+D_(n-2)`

B

`D_(n-1)+(n-1)D_(n-2)`

C

`n(D_(n-1)+D_(n-2))`

D

`(n-1)(D_(n-1)+D_(n-2))`

Text Solution

Verified by Experts

The correct Answer is:
D
`(d)` For every choice of `r=1,2,3,….(n-1)` when the `n^(th)` object `a_(n)` goes to the `rth` place, there are `D_(n-1)+D_(n-2)` ways of the other `(n-1)` objects `a_(1)`, `a_(2)`, ….,`a_(n-1)` to be deranged.
Hence `=D_(n)=(n-1)(D_(n-1)+D_(n-2))`
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