Question
If $${a_1},{a_2},{a_3},.....$$ are in A.P. and $$a_1^2 - a_2^2 + a_3^2 - a_4^2 + ..... + a_{2k - 1}^2 - a_{2k}^2 = M\left( {a_1^2 - a_{2k}^2} \right).\,{\text{Then }}M = $$
A.
$$\frac{{k - 1}}{{k + 1}}$$
B.
$$\frac{{k }}{{2k - 1}}$$
C.
$$\frac{{k + 1}}{{2k + 1}}$$
D.
none
Answer :
$$\frac{{k }}{{2k - 1}}$$
Solution :
We have, $${a_2} - {a_1} = {a_3} - {a_2} = .....\,{a_{2k}} - {a_{2k - 1}} = d$$
Hence,
$$\eqalign{
& a_1^2 - a_2^2 = \left( {{a_1} - {a_2}} \right)\left( {{a_1} + {a_2}} \right) = - d\left( {{a_1} + {a_2}} \right) \cr
& a_3^2 - a_4^2 = \left( {{a_3} - {a_4}} \right)\left( {{a_3} + {a_4}} \right) = - d\left( {{a_3} + {a_4}} \right) \cr
& ......................... \cr
& ......................... \cr
& a_{2k - 1}^2 - a_{2k}^2 = \left( {{a_{2k - 1}} - {a_{2k}}} \right)\left( {{a_{2k - 1}} + {a_{2k}}} \right) = - d\left( {{a_{2k - 1}} + {a_{2k}}} \right) \cr
& {\text{Adding, we get}} \cr
& a_1^2 - a_2^2 + a_3^2 - a_4^2 + ..... + a_{2k - 1}^2 - a_{2k}^2 \cr
& = - d\left( {{a_1} + {a_2} + {a_3} + {a_4} + .....\,{a_{2k - 1}} + {a_{2k}}} \right) \cr
& = - d \cdot \frac{{2k}}{2}\left( {{a_1} + {a_{2k}}} \right) = - dk\left( {{a_1} + {a_{2k}}} \right) \cr
& {\text{But, }}{a_{2k}} = {a_1} + \left( {2k - 1} \right)d \cr
& \Rightarrow - d = \frac{{{a_1} - {a_{2k}}}}{{2k - 1}} \cr
& \therefore {\text{The required sum}} = \frac{k}{{2k - 1}}\left( {a_1^2 - a_{2k}^2} \right) \cr
& \Rightarrow M = \frac{k}{{2k - 1}} \cr} $$