Question

If ${\text{17}}^{th}$ term of an arithmetic progression exceeds its ${10}^{\mathrm{th}}$ term by $7$. Find the common difference.

Answer 1

Step 1; Define the terms of the arithmetic progression in standard form

The ${\mathrm{n}}^{\mathrm{th}}$ term of an arithmetic progression is given as

${\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}$

where $\mathrm{a}$is the first term and $\mathrm{d}$is the common difference

${\mathrm{a}}_{17}=\mathrm{a}+(17-1)\mathrm{d}$

${\mathrm{a}}_{17}=\mathrm{a}+16\mathrm{d}$ $...\left(\mathrm{i}\right)$

${\mathrm{a}}_{10}=\mathrm{a}+(10-1)\mathrm{d}$

${\mathrm{a}}_{10}=\mathrm{a}+9\mathrm{d}$ $...\left(\mathrm{ii}\right)$

Step 2: Use the given relation to find the common difference

From the given relation we know

${\mathrm{a}}_{17}-{\mathrm{a}}_{10}=7$

From $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$ we get

$(\mathrm{a}+16\mathrm{d})-(\mathrm{a}+9\mathrm{d})=7$

$\Rightarrow$ $7\mathrm{d}=7$

$\Rightarrow$ $\mathrm{d}=1$

Hence, the common difference is $1$.