Question

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

Answer 1

     Given is a17 i.e a + (n-1)d = a + 16d

      Given is a10 i.e a + (n-1) d = a + 9d 

Given that 17th term exceeds its 10th term by 7, i.e
 
a + 16d  = a + 9d + 7

On solving we get,

a + 16d = a + 9d + 7

= 16d - 9d = 7

7d = 7

d = 1

Hope this helps!

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.