Question

Find the sum of first 11 positive numbers which are multiple of 6.

Answer 1

The first 11 positive numbers that are multiples of 6 are:
6, 12, 18,…, 60, 66
In this A.P., we have:
a = 6 and d = t₂ – t₁ = 12 – 6 = 6
We know:
Sum of n terms of an A.P., $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

∴ Sum of the first 11 positive numbers that are multiples of 6:

$$\begin{gathered} S_{11}=\frac{11}{2}\left[2\times 6+\left(11-1\right)\times 6\right] \\ =\frac{11}{2}\left[12+10\times 6\right] \\ =\frac{11}{2}\left[12+60\right] \\ =\frac{11}{2}\times 72 \\ =11\times 36=396 \end{gathered}$$