Question

The 17th term of an AP exceeds it's 10th term by 7.Find the common difference​

Answer 1

Answer:

Given that 17th term of an AP exceeds its 10th term by 7.

i.e.

$a_{17} = a_{10} + 7$

By putting this condition we can easily find the value of common difference .

Let,

First term of the AP = a

and

common difference of AP = d

$\cancel a + 16d = \cancel a+ 9d + 7 \\ \\16d - 9d = 7 \\ \\ 7d = 7 \\ \\ d = 1$

So,

common difference of the AP is 7

Answer 2

Answer

a+(n-1)d

Therefore : a17 =a+(17-1)d

a17=a+16d.............(1)

Then : a10= a+(10-1)d

a10=a+9d...........(2)

So: a17-a10

(a+16d)-(a+9d)=7

a+16d-a-9d=7

collect like terms together

a-a+16d-9d=7

: 0+7d=7

7d=7

Divide both sides by 7

Common difference (d)= 1