Please solve question no.iv.

Dear Student

The list of natural numbers divisible by 6 are :
6, 12, 18, ...........96
The above list forms an AP, with a = 6  ,  d = 6
Let a_n = 96   , Let n = no. of terms
$$\begin{gathered} a_n = a + (n - 1)d \\ \Rightarrow 96 = 6 + (n - 1)6 \\ \Rightarrow 90 = 6n - 6 \\ \Rightarrow 6n = 96 \\ \Rightarrow n = 16 \\ Now, \\ {S}_n = \frac{n}{2}\left[2a + (n - 1)d\right] \\ so, S_{16} = \frac{16}{2} \left[2\left(6\right) + (16 - 1) 6\right] = 8 [12 + 90] = 8 \times 102 = 816 \end{gathered}$$

Regards