# If a1,a2, a3, ........ , an and b1, b2, b3, ........ ,bn are two Arithmetic Progressions.Is a1 + b1, a2 + b2, ........, an + bn an Arithmetic Progression? Why/ Why not?

2
by raokarthik19

2015-03-23T11:16:42+05:30

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Let a1 be 1
a2 be 2
a3 be 3 and so on
so the arithemetic progression will be 1,2,3,4,....
let b1 be 2
b2 be 2
b3 be 2 and so on
so the arithemetic progression will be 2,2,2,2,...
a1+ b1 = 1+2 =3
a2+ b2 = 2+2 =4
a3 + b3 = 3+2 =5
a4 + b4 =4+2 =6 and so on
so the arithemietic progression will be 3,4,5,6....
to verify it 4-3 should be equal to 5-4 which is equal to 1
so it is an arithemetic progression
2015-03-24T01:19:32+05:30

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Yes, (a1+b1), (a2+b2), (a3+b3),........,(an+bn) is in AP

Let Sequence a1, a2, a3,....an have common difference be (d1)
And that of b1, b2, b3,....bn be d2

According to condition:-
Sequence (a1+b1), (a2+b2), (a3+b3),........,(an+bn),
Now the each succeding terms of this sequence differ by d1+ d2 which is constant for all the terms of this AP. So it is an AP.

For eq:-
Let A = a1, a2, a3,....an = 2,4,6,8....(here d = 2)
Let B = b1, b2, b3,....bn = 1,3,5,7,....(here also d = 2)
Now sequence (a1+b1), (a2+b2), (a3+b3),........,(an+bn) = (2+1), (4+3), (6+5), (8+7) = 3,7,11,15......
In this sequence, 7-3=4, 11-7 = 4, 15-11 = 4. So it is an AP.....!