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# Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7 . The difference between their $ 10^{\text {th }} $ terms is the same as the difference between their $ 21^{\text {st }} $ terms, which is the same as the difference between any two corresponding terms. Why?

Given:

Two A.P.s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms.

To do:

We have to find the reason behind the given statement.

Solution:

Let $a, a+d, a+2d,......$ and $p, p+d, p+2d,.......$ be the two A.P.s.

Therefore,

$a=2$ and $p=7$

$a_{10}=a+(10-1)d$

$=a+9d$

$p_{10}=p+(10-1)d$

$=p+9d$

$a_{21}=a+(21-1)d$

$=a+20d$

$p_{21}=p+(21-1)d$

$=p+20d$

Therefore,

$a_{10}-p_{10}=a+9d-(p+9d)$

$=a-p$

$a_{21}-p_{21}=a+20d-(p+20d)$

$=a-p$

As we can see, when the common difference of any two A.P.s is equal, the difference between any two corresponding terms is always equal to the difference of the first terms of the two A.P.s.

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