Let $${A_n}$$ be the sum of the first $$n$$ terms of the geometric series $$704 + \frac{{704}}{2} + \frac{{704}}{4} + \frac{{704}}{8} + ....$$       and $${B_n}$$ be the sum of the first $$n$$ terms of the geometric series $$1984 - \frac{{1984}}{2} + \frac{{1984}}{4} + \frac{{1984}}{8} + ....$$       If $${A_n} = {B_n},$$   then the value of $$n$$ is (where $$n \in N$$  ).

Solution :
$$\eqalign{ & {A_n} = 704 + \frac{{704}}{2} + \frac{{704}}{4} + ....\,{\text{to }}n{\text{ terms}} \cr & = \frac{{704\left( {1 - {{\left( {\frac{1}{2}} \right)}^n}} \right)}}{{1 - \frac{1}{2}}} \cr & = 704 \times 2\left( {1 - {{\left( {\frac{1}{2}} \right)}^n}} \right) \cr & {B_n} = 1984 - \frac{{1984}}{2} + \frac{{1984}}{4} + ....\,{\text{to }}n{\text{ terms}} \cr & = \frac{{1984\left( {1 - {{\left( {\frac{{ - 1}}{2}} \right)}^n}} \right)}}{{1 - \left( {\frac{{ - 1}}{2}} \right)}} \cr & = 1984 \times \frac{2}{3}\left( {1 - {{\left( {\frac{{ - 1}}{2}} \right)}^n}} \right) \cr & {\text{Now}},\,{A_n} = {B_n} \cr & \Rightarrow 704 \times 2\left( {1 - {{\left( {\frac{1}{2}} \right)}^n}} \right) \cr & = 1984 \times \frac{2}{3} \times \left( {1 - {{\left( {\frac{{ - 1}}{2}} \right)}^n}} \right) \cr & \Rightarrow 33 - 31 = 33{\left( {\frac{1}{2}} \right)^n} - 31{\left( {\frac{{ - 1}}{2}} \right)^n} \cr & \Rightarrow {2^{n + 1}} = 33 - 31{\left( { - 1} \right)^n} \cr & \Rightarrow n = 5 \cr} $$