• # question_answer 9) Find the sum of first 8 multiples of 3.

 First 8 multiples of 3 are 3, 6, 9............. up to 8 terms We can observe that the above series is an AP with $a=3,\text{ }d=6-3=3,n=8$ Sum of n terms of an A.P. is given by, ${{S}_{n}}=\frac{n}{2}[2a+(n-1)d]$ $\therefore$      ${{S}_{8}}=\frac{8}{2}[2\times 3+(8-1)(3)]$ $=4[6+7\times 3]$ $=4[6+21]$ $=4\times 27$ $\Rightarrow$   ${{S}_{8}}=108$