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Question

# Let a1,a2,a3,..........andb1b2,b3........... be arithmetic progression such that a1=25,b1=75anda100+b100=100, then

A
The common difference in progression ai is equal but opposite in sign to the common difference in progression bj.
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B
an+bn = 100 for any n.
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C
(a1+b1),(a2+b2),(a3+b3), .......... are in A.P.
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D
100r=1(ar+br)=104
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Solution

## The correct options are A The common difference in progression ′a′i is equal but opposite in sign to the common difference in progression ′b′j. B an+bn = 100 for any n. C 100∑r=1(ar+br)=104 D (a1+b1),(a2+b2),(a3+b3), .......... are in A.P.Given a1,a2,a3,.......... and b1b2,b3........... be two AP's.Let d1,d2 be the common difference of the two AP's.Then, the AP's can be written as a1,a1+d1,a1+2d1..........b1,b1+d2,b1+2d2..........Now, given a100+b100=100⇒a1+99d1+b1+99d2=100⇒100+99(d1+d2)=100⇒d1+d2=0⇒d1=−d2Hence, option A is correct.Now, an+bn=an+(n−1)d1+b1+(n−1)d2=a1+b1=100Hence, option B is correct.Since, an+bn=100∀nSo, a1+b1,a2+b2+.... are in AP.Hence, option C is also true.Now 100∑r=1(ar+br))=a1+b1,a2+b2+....+a99+b99+a100+b100=100+100+....upto100times=104Hence, option D is also true.

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