JEE Main & Advanced Mathematics Sequence & Series Question Bank Self Evaluation Test - Sequences and Series

If \[{{a}_{1}},{{a}_{2}},{{a}_{3}}.....\] are in A.P. and \[a_{1}^{2}-a_{2}^{2}+a_{3}^{2}-a_{4}^{2}+......+a_{2k-1}^{2}-a_{2k}^{2}\] \[=M(a_{1}^{2}-a_{2k}^{2}).\] Then M =

A) \[\frac{k-1}{k+1}\]

B) \[\frac{k}{2k-1}\]

C) \[\frac{k+1}{2k+1}\]

D) None

Correct Answer: B

Solution :

[b] We have, \[{{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}\]

\[=..............{{a}_{2k}}-{{a}_{2k-1}}=d\]

Hence,

\[a_{1}^{2}-a_{2}^{2}=({{a}_{1}}-{{a}_{2}})\,({{a}_{1}}+{{a}_{2}})=-d({{a}_{1}}+{{a}_{2}})\]

\[a_{3}^{2}-a_{4}^{2}=({{a}_{3}}-{{a}_{4}})({{a}_{3}}+{{a}_{4}})=-d({{a}_{3}}+{{a}_{4}})\]

??????..

\[a_{2k-1}^{2}-a_{2k}^{2}=({{a}_{2k-1}}-{{a}_{2k}})({{a}_{2k-1}}+{{a}_{2k}})\]

\[=-d({{a}_{2k-1}}+{{a}_{2k}})\]

Adding, we get

\[a_{1}^{2}-a_{2}^{2}+a_{3}^{2}-a_{4}^{2}+............+a_{2k-1}^{2}-a_{2k}^{2}\]

\[=-d({{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+..........{{a}_{2k-1}}+{{a}_{2k}})\]

\[=\,\,\,-\,\,d\,.\,\frac{2k}{2}({{a}_{1}}+{{a}_{2k}})=-dk({{a}_{1}}+{{a}_{2k}})\]

But \[{{a}_{2k}}={{a}_{1}}+(2k-1)d\Rightarrow -d=\frac{{{a}_{1}}-{{a}_{2k}}}{2k-1}\]

\[\therefore \] The required sum \[=\frac{k}{2k-1}\left( a_{1}^{2}-a_{2k}^{2} \right)\]

\[\Rightarrow \,\,M=\frac{k}{2k-1}\]