Answer: Option D. -> A is larger than B by 1
B is in G.P. with a = 20, r = 2, n = 65
$$\eqalign{
& \therefore {S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{2^0}\left( {{2^{65}} - 1} \right)}}{{2 - 1}} \cr
& \therefore B = {2^{65}} - 1 \cr
& \Rightarrow B = A - 1 \cr
& \therefore A\,{\text{is}}\,{\text{larger}}\,{\text{then}}\,B\,{\text{by}}\,1 \cr} $$

Answer: Option A. -> $$\frac{{16}}{{81}}$$
Each time the ball is dropped and it bounces back, it reaches $$\frac{{2}}{{3}}$$ of the height it was dropped from.
After the first bounce, the ball will reach $$\frac{{2}}{{3}}$$ of the height from which it was dropped - let us call it the original height.
After the second bounce, the ball will reach $$\frac{{2}}{{3}}$$ of the height it would have reached after the first bounce.
So, at the end of the second bounce, the ball would have reached $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ of the original height = $$\frac{{4}}{{9}}$$ th of the original height.
After the third bounce, the ball will reach $$\frac{{2}}{{3}}$$ of the height it would have reached after the second bounce.
So, at the end of the third bounce, the ball would have reached $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ = $$\frac{{8}}{{27}}$$ th of the original height.
After the fourth and last bounce, the ball will reach $$\frac{{2}}{{3}}$$ of the height it would have reached after the third bounce.
So, at the end of the last bounce, the ball would have reached $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ of the original height = $$\frac{{16}}{{81}}$$ of the original height.