Question

the 17th term of an ap exceeds its 10th term by 7 find the common difference

Answer 1

Term 17 = a + 16d
Term 10 = a + 9d
Given that   Term 17 - Term 10 = 7
⇒ a + 16d - (a + 9d) = 7
⇒a + 16d - a - 9d = 7
⇒7d = 7
⇒d = 7/7
∴The common difference     d = 1

Answer 2

Solution :

Let a be the first term and d the common difference.

It is given       $a_{17} - a_{10} = 7$

⇒ (a + 16d) - (a + 9d) = 7

⇒                          7d = 7  

⇒                            d = 1

Thus, the common difference is 7.