Question

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

Answer 1

We know that, For an A.P.,
$a_n = a + (n − 1) d$
$a_{17} = a + (17 − 1) d$
$a_{17} = a + 16d$ ……. (i)
Similarly,
$a_{10} = a + (10− 1) d$
$a_{10}= a + 9d$ ……. (ii)
It is given that,
$a_{17} − a_{10} = 7$
$(a + 16d) − (a + 9d) = 7$
$7d = 7$
$d = 1$

Therefore, the common difference is 1.