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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