Question

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

Answer 1

Answer:

Let a be the first term and d be the common difference of A.P.

17th term of A.P t17 = a + 16d.

10th term of A.P is t10 = a + 9d.

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

a + 16d = a + 9d + 7 7d = 7

∴ d = 1

Answer 2

Given

A word problem

To find

Common difference =cd

Solution

According to the question

17th term = 10th term + 7 ( In an easy language)

we have to find 17th and 10th term

$\implies{T_n}=a+(n-1)d}$

Now $\implies{T_{10}}$

$\implies{T_{10}=a+(10-1)d}$

$\implies{T_{10}=a+(9)d}$

$\implies{T_{19}=a+9d}$

Similarly $\implies{T_17}$

$\implies{T_{17}=a+(17-1)d}$

$\implies{T_{17}=a+16d}$

Putting the values

=>a+16d=a+9d+7

=>16d-9d=7

=>7d=7

d=1

Hence,the value of common difference be 1