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
`1000`
B
`100000`
C
`10000`
D
`24000`
Text Solution
Verified by Experts
The correct Answer is:
C
`(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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CENGAGE-PROGRESSION AND SERIES-ARCHIVES (MATRIX MATCH TYPE )
