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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 ai is equal but opposite in sign to the common difference in progression bj.
B an+bn = 100 for any n.
C 100r=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=d2
Hence, option A is correct.
Now, an+bn
=an+(n1)d1+b1+(n1)d2
=a1+b1=100
Hence, option B is correct.
Since, an+bn=100n
So, a1+b1,a2+b2+.... are in AP.
Hence, option C is also true.
Now 100r=1(ar+br))
=a1+b1,a2+b2+....+a99+b99+a100+b100
=100+100+....upto100times
=104
Hence, option D is also true.

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