Question

In an AP ${17}^{th}$ term is $7$ more than its ${10}^{th}$ term. Find the common difference.

Answer 1

Find the common difference

Given: In an AP ${17}^{th}$ term is $7$ more than its ${10}^{th}$ term.

We know that the $n^{th}$ term of an AP is calculated by using the formula, $a_n=a+\left(n-1\right)d$.

So, we can say that the ${10}^{th}$term can be written as, $a_{10}=a+\left(10-1\right)d=a+9d$ and we can write the ${17}^{th}$ term as,$a_{17}=a+\left(17-1\right)d=a+16d$.
Now, according to the question,
$a_{17}=7+a_{10}$
By substituting the values of $a_{17}$ and $a_{10}$ in the above equation. we get,
$$\begin{gathered} a+16d=7+a+9d \\ \Rightarrow 16d-9d=7+a-a \\ \Rightarrow 7d=7 \\ \Rightarrow d=\frac{7}{7} \\ \Rightarrow d=1 \end{gathered}$$
Hence, the common difference of the given AP is $1$.