Let $${a_1},{a_2},.....,{a_{30}}$$    be an A.P., $$S = \sum\limits_{i = 1}^{30} {{a_i}} {\text{ and }}T = \sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}}.} $$      If $${a_5} = 27{\text{ and }}S - 2T = 75,$$     then $${a_{10}}$$ is equal to:

Solution :
$$\eqalign{ & S = \sum\limits_{i = 1}^{30} {{a_i} = \frac{{30}}{2}\left[ {2{a_1} + 29d} \right]} {\text{ }} \cr & T = \sum\limits_{i = 1}^{15} {{a_{\left( {2i - 1} \right)}} = \frac{{15}}{2}\left[ {2{a_1} + 28d} \right]} \cr & {\text{Since, }}S - 2T = 75 \cr & \Rightarrow \,\,30\,{a_1} + 435d - 30\,{a_1} - 420d = 75 \cr & \Rightarrow \,\,d = 5 \cr & {\text{Also, }}{a_5} = 27\,\,\,\, \Rightarrow \,\,{a_1} + 4d = 27 \cr & \Rightarrow \,\,{a_1} = 7, \cr & {\text{Hence, }}{a_{10}} = {a_1} + 9d = 7 + 9 \times 5 = 52 \cr} $$