Question

The sum of all two-digit natural numbers which leave a remainder $5$ when they are divided by $7$ is equal to

• A. $715$
• B. $702$
• C. $615$
• D. $602$

Answer 1

The correct option is B

$702$

Explanation for correct option:

The two digit number which leaves remainder $5$, when divided by $7$ are $12,19,26,....,89,96$

Here the first term $a_1=12$

Second term $a_2=19$

Therefore common difference, $d=a_2-a_1=19-12=7$

Let the total number of terms $n$

Last term $a_n=96$

We know that last term $a_n=a_1+d\left(n-1\right)$

$$\begin{gathered} \Rightarrow a_1+d\left(n-1\right)=96 \\ \Rightarrow 12+7\left(n-1\right)=96 \\ \Rightarrow n-1=\frac{96-12}{7} \\ \Rightarrow n=13 \end{gathered}$$

We know that Sum of $n$ terms of AP$=\frac{n}{2}\left(a_1+a_n\right)$
$$\begin{gathered} =\frac{13}{2}\left(12+96\right) \\ =\frac{13}{2}\left(108\right) \\ =13\times 54 \\ =702 \end{gathered}$$

Hence, option B is correct.