JEE MAIN - Mathematics (2006 - No. 24)
Let $${a_1}$$, $${a_2}$$, $${a_3}$$.....be terms on A.P. If $${{{a_1} + {a_2} + .....{a_p}} \over {{a_1} + {a_2} + .....{a_q}}} = {{{p^2}} \over {{q^2}}},\,p \ne q,\,then\,{{{a_6}} \over {{a_{21}}}}\,$$ equals
$${{41} \over {11}}$$
$${7 \over 2}$$
$${2 \over 7}$$
$${{11} \over {41}}$$
Explanation
$${{{p \over 2}\left[ {2{a_1} + \left( {p - 1} \right)d} \right]} \over {{q \over 2}\left[ {2{a_1} + \left( {q - 1} \right)d} \right]}} = {{{p^2}} \over {{q^2}}}$$
$$ \Rightarrow {{2{a_1} + \left( {p - 1} \right)d} \over {2{a_1} + \left( {p - 1} \right)d}} = {p \over q}$$
$${{{a_1} + \left( {{{p - 1} \over 2}} \right)d} \over {{a_1} + \left( {{{q - 1} \over 2}} \right)d}} = {p \over q}$$
For $${{{a_6}} \over {a{}_{21}}},\,\,p = 11,\,q = 41$$
$$ \Rightarrow {{{a_6}} \over {a{}_{21}}} = {{11} \over {41}}$$
$$ \Rightarrow {{2{a_1} + \left( {p - 1} \right)d} \over {2{a_1} + \left( {p - 1} \right)d}} = {p \over q}$$
$${{{a_1} + \left( {{{p - 1} \over 2}} \right)d} \over {{a_1} + \left( {{{q - 1} \over 2}} \right)d}} = {p \over q}$$
For $${{{a_6}} \over {a{}_{21}}},\,\,p = 11,\,q = 41$$
$$ \Rightarrow {{{a_6}} \over {a{}_{21}}} = {{11} \over {41}}$$
Comments (0)
