Question

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

Answer 1

a/q 
a17=a+16d
a10=a+9d
a/q
a17=a10+7
a+16d=a+9d+7     [a will be canceled]
16d-9d=7
7d=7
d=1

Answer 2

☺ Hello mate__ ❤

◾◾here is your answer...

Given, 17th term exceeds its 10th term by 7.

It means a17 = a10 + 7         ........eq(1)

Here, a17 is the 17th term and a10 is the 10th term of an AP.

Using formula an=a+(n−1)d,   to find nth term of arithmetic progression, we get

a17=a+(16)d      ..........eq (2)

a10=a+(9)d    .........eq(3)

Putting (2) and (3) in equation (1), we get

a+16d=a+9d+7

⇒7d=7

⇒d=7/7=1

I hope, this will help you.

Thank you______❤

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