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

`1000`

`100000`

`10000`

`24000`

Text Solution

Verified by Experts

The correct Answer is:

`(c )` `(a_(100)+b_(100))-(a_(1)+b_(1))=99dimpliesd=0`
`:.` Sum of series `=(100//2)(a_(100)+b_(100)+a_(1)+b_(1))`
`=50(100+100)`
`=10000`

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