Question

Find the sum of all two-digit numbers greater than $50$, which when divided by $7$ leaves remainder $4$.

Answer 1

Step 1: Find the first term, common difference and $\mathit{n}$:

Formula:

The $n^{\mathrm{th}}$ term of an AP is $a_n=a+\left(n-1\right)d$.

According to the question the required AP sequence is $53,60,67,\dots ,95$.

Here, the first term, $a=53$, the common difference, $d=60-53$, that is, $d=7$ and the $n^{\mathrm{th}}$ term is, $a_n=95$. Substitute these values in the $n^{\mathrm{th}}$ term of an AP formula.

$$\begin{gathered} \Rightarrow 95=53+\left(n-1\right)7 \\ \Rightarrow 42=7n-7 \\ \Rightarrow 7n=49 \\ \Rightarrow n=7 \end{gathered}$$

Step 2: Evaluate the sum of $\mathbf{7}$ terms:

Formula:

The sum of $n$ terms of a series in an AP is $S_n=\frac{n}{2}\left[a+a_n\right]$.

Substitute $a=53$, $a_n=95$ and $n=7$ into the $S_n$ formula.

$\begin{array}{rcl}& \Rightarrow & S_7=\frac{7}{2}\left(53+95\right)\\ & =& \frac{7}{2}\times 148\\ & =& 518\end{array}$

Hence, the required sum is $518$.