Question

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

Answer 1

It is given that ${17}^{th}$ term exceeds its ${10}^{th}$ term by $7$

It means $a_{17}$ $=a_{10}+\mathrm{7......}\left(1\right)$

Here, $a_{17}$ is the ${17}^{th}$ term and a₁₀ is the ${10}^{th}$ term of an AP.

Using formula $a_n=a+(n-1)d$ to find $n^{th}$ term of arithmetic progression.

Putting $n=17$ and $n=10$ in the above formula to find $a_{17}$ and $a_{10}$ respectively.

$a_{17}$$=a+\left(16\right)d.....\left(2\right)$

$a_{10}$$=a+\left(9\right)d.....\left(3\right)$

Putting eq. $\left(2\right)$ and $\left(3\right)$ in equation $\left(1\right),$ we get

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

$\Rightarrow 7d=7$

$\Rightarrow d=\frac{7}{7}=1$