Question 17

16, 32, 64, 128, ....The sum of first 10 numbers in the series is:

Solution

$$16+16\times\ 2+16\times\ 4+16\times\ 8,,,,,,$$

16($$1+\ 2+\ 4+\ 8,,,,,,$$),

it's sum of G.P so $$a\left(\ \frac{\ r^n-1}{r-1}\right)$$, a=1, r=2, n=10.

16$$\times\ $$($$1\times\ \ \frac{\ 2^{10}-1}{2-1}$$)=16368

hence option "A" is correct.

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