Question

6. Find the sum of all the numbers less than $100$ which are divisible by $6$.

Answer 1

Step 1: Note the given data

Given the series is the numbers less than $100$ which are divisible by $6$.

The first digit in the natural number is $a=6$ and the last term is $96$, which are divisible by $6$

Therefore the numbers are $6,12,18,.......,96$

Let the last term $t_n=96$

Step 2: Find the common difference

The general form for finding the common difference $d={(n+1)}^{th}term-n^{th}term$

Common difference

$$\begin{gathered} d=12-6 \\ =6 \end{gathered}$$

Similarly, $d=12-6=18-12=6$

So, the series are in A.P.

Step 3: Find the $n^{th}$ term of A.P.

The formula for finding $n^{th}$term of A.P. $=a+(n-1)d$

The value of $n^{th}$term of A.P $=6+(n-1)\times 6$

$$\begin{gathered} =6+6n-6 \\ =6n \end{gathered}$$

Step 4: Equating both the $n^{th}$ value of A.P.

$6n=96$

$$\begin{gathered} \Rightarrow n=\frac{96}{6} \\ \Rightarrow n=16 \end{gathered}$$

Step 5: Find the required sum of the given series

The formula of the sum of the series of A.P. $=\frac{n}{2}(t_1+t_n)$

Sum of the given series

$$\begin{gathered} =\frac{16}{2}(6+96) \\ =8\times 102 \\ =816 \end{gathered}$$

Hence, the sum of the given series is $816$.