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Question

The $$17^{th}$$ term of an $$AP$$ exceeds its $$10^{th}$$ term by $$7$$. Find the common difference.


Solution

We know that,

For an A.P $$n^{th}$$ term is given as $$a_{n}=a+(n-1)d$$
$$a_{17}=a+(17-1)d$$
$$a_{17}=a+16d$$

Similarly, $$a_{10}=a+9d$$

It is given that
$$a_{17}-a_{10}=7$$
$$(a+16d)-(a+9d)=7$$
$$7d=7$$
$$\therefore d=1$$

Therefore, the common difference is $$1$$.

Mathematics
RS Agarwal
Standard X

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