Question

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

Answer 1

Given:

${a}_{17} = {a}_{10} + 7$ (Equation 1)

a = Term in series

d = common difference

Solution:

${a}_{17} = a + 16d$ (Equation 2)

${a}_{10} = a + 9d$ (Equation 3)

So,

From Equation 1, Equation 2 and Equation 3

a + 16d = a + 9d + 7

16d - 9d = 7

7d = 7

d = 1

Hence, common difference (d) = 1

Answer 2

Step-by-step explanation:

So, we know that formula of A.P. is

$a_n=a+(n-1)d$

where,

a=the first term of the A.P.

d=the common difference of the A.P.

Now, lets come to our question,

The expression for 17th term will be given by,

17th term=a+(17-1)d

a+16d --(1)

Now, the expression for 10th term,

10th term=a+(10-1)d

a+9d --(2)

According to question,

The 17th term exceeds the 10th term by 7, so

a+16d=a+9d+7 --(3)

16d-9d=7

16d-9d=77d=7

therefore, d=1

Hence, the common difference is 1.