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 \[=\frac{k}{2k-1}\left( a_{1}^{2}-a_{2k}^{2} \right)\] |
\[\Rightarrow \,\,M=\frac{k}{2k-1}\] |
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