1.

Find the sum of 8 terms of the GP 3, 6, 12, 24, ...

Answer» Here `a=3, r=2 gt 1` and `n=8`.
Using the formula, `S_(n)=(a(r^(n)-1))/((r-1))`, we get
`S_(8)=(3xx(2^(8)-1))/((2-1))=3xx(256-1)=3xx255=765`.


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