Sequence And Series Question 506
Question: Let the sequence $ a_1,a_2,a_3,………….a_{2n} $ form an A.P. Then $ a_1^{2}-a_2^{2}+a_3^{3}-………+a_{2n-1}^{2}-a_2n^{2}= $
Options:
A) $ \frac{n}{2n-1}(a_1^{2}-a_2n^{2}) $
B) $ \frac{2n}{n-1}(a_2n^{2}-a_1^{2}) $
C) $ \frac{n}{n+1}(a_1^{2}+a_2n^{2}) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Since $ a_1,\ a_2,,a_3,………..,a_{n} $ form an A.P. therefore, $ a_2-a_1=a_4-a_3=…….=a_{2n}-{a_{2n-1}}=d $ Here $ a_1^{2}-a_2^{2}+a_3^{2}-a_4^{2}+…….+a_{2n-1}^{2}-a_2n^{2} $ $ =(a_1-a_2)(a_1+a_2)+(a_3-a_4)(a_3+a_4)+…….. $ $ ……+({a_{2n-1}}-a_{2n})({a_{2n-1}}+a_{2n}) $ $ =-d(a_1+a_2+…….+a_{2n})=-d{ \frac{2n}{2}(a_1+a_{2n}) } $ Also we know $ a_{2n}=a_1+(2n-1)d $
$ \Rightarrow $ $ d=\frac{a_{2n}-a_1}{2n-1} $
$ \Rightarrow $ $ -d=\frac{a_1-a_{2n}}{2n-1} $ .
$ \therefore $ Therefore the sum is = $ \frac{n(a_1-a_{2n}).(a_1+a_{2n})}{2n-1}=\frac{n}{2n-1}(a_1^{2}-a_2n^{2}) $ .
BETA
