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# The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Given:

The 17th term of an AP exceeds its 10th term by 7.

To do:

We have to find the common difference.

Solution:

Let the first term and the common difference of the given A.P. be $a$ and 4$d$ respectively.

We know that,

nth term of an A.P. $a_n=a+(n-1)d$

Therefore,

$a_{10}=a+(10-1)d=a+9d$

$a_{17}=a+(17-1)d=a+16d$

According to the question,

$a_{17}=a_{10}+7$

$a+16d=a+9d+7$

$16d-9d=7$

$7d=7$

$d=1$

The common difference is $1$.

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