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

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#### Solution

We know that,

For an A.P., 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

`d = 7/7`

d = 1

Therefore, the common difference is 1

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