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

Q : 10      The $\small 17$th term of an AP exceeds its $\small 10$th term by $\small 7$. Find the common difference.

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It is given that
$\small 17$th term of an AP exceeds its $\small 10$th term by $\small 7$
i.e.
$a_{17}= a_{10}+7$
$\Rightarrow a+16d = a+9d+7$
$\Rightarrow a+16d - a-9d=7$
$\Rightarrow 7d=7$
$\Rightarrow d = 1$
Therefore, the common difference of AP is 1

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