Answer:
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, \[{{S}_{n}}=\frac{n}{2}[2a+(n-1)d]\] \[\therefore \] \[{{S}_{8}}=\frac{8}{2}[2\times 3+(8-1)(3)]\] \[=4[6+7\times 3]\] \[=4[6+21]\] \[=4\times 27\] \[\Rightarrow \] \[{{S}_{8}}=108\]
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