Sequence And Series Question 393
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(a_1^{2}-a_2k^{2}). $ Then M =
Options:
A) $ \frac{k-1}{k+1} $
B) $ \frac{k}{2k-1} $
C) $ \frac{k+1}{2k+1} $
D) None
Show Answer
Answer:
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}( a_1^{2}-a_2k^{2} ) $
$ \Rightarrow M=\frac{k}{2k-1} $
BETA
