When the corresponding terms of two AP are added are the terms of the resulting series also an AP?

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william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

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To determine whether the corresponding terms of two AP when added yield the terms of a new AP, let’s take two APs.

In one the first term is a1 and the common difference is d1, in the other the first term is a2 and the common difference is d2.

Now for the first AP the nth term is a1 + (n-1) d1.

For the second AP the nth term is a2 + (n-1) d2

Now add the two terms:   a1 + (n-1) d1 + a2 + (n-1) d2 = a1+ a2 + (n-1) (d1+ d2)

This is a new AP with the first term as a1+ a2 and the common difference d1+ d2.

Hence we have got a new AP.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Yes.

If two series which are in arithmetic progression are added, term by term  correspondingly, then the resulting  series is also in AP.

Proof:

  Let  a1 ,  d, and n be the first term , the common difference and the number of terms  in a series series A  of an AP.

Then the nth term an of the series A  is giveny by the relation:

 an = a1+(n-1)d

Similarly let , b1 , e  and n be the first term, the common difference  and the number of the terms of another series B.

Then the nth term bn of the series B is given by: b1+(n-1)e.

Clearly , during the the addition of the corresponding terms nth terms of the series A and B is : an+bn = a1+(n-1)d }+{(b1+(n-1)e)} = (a1+b1) + (n-1)(d+e). Or

The nth term  ahter adding the corresponding term bu terms of both series is given by (an+bn) = (a1+b1) +(n-1)(d+e)  = k1+(n-1)f , where f = d+e.

Thus the new series is  also an AP with starting term  a1+b1 = k1 and a common ratio d+e = f.

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