Question

The ${17}^{th}$ term of an $AP$ exceeds its ${10}^{th}$ term by $7$. Find the common difference.

Answer 1

We know that,

For an A.P $n^{th}$ term is given as $a_n=a+(n-1)d$

$a_{17}=a+(17-1)d$

$a_{17}=a+16d$

Similarly, $a_{10}=a+9d$

It is given that

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

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

$7d=7$

$\therefore d=1$

Therefore, the common difference is $1$.