Quantitative Aptitude Ques 2403

Question: What is the value of $ [1^{2}-2^{2}+3^{3}-4^{2}+5^{2}-6^{2}+…+11^{2}-12^{2}]? $

Options:

A) 55

B) $ -78 $

C) $ -,55 $

D) 78

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ [1^{2}-2^{2}+3^{2}-4^{2}+5^{2}-6^{2}+….+11^{2}-12^{2}] $ $ =[(1+2)(1-2)+(3+4)(3-4)+(5+6) $ $ (5-6)+…+(11+12)(11-12)] $ $ =[3\times -1+7\times -1+11\times -1+…+23\times -1] $ $ =[-,3-7-11-….-23] $ This series is in AP with first term $ =-,3 $ and last term $ =-,23 $ $ \because $ $ l=a+(n-1),d $ $ -,23=-,3+(n-1)\times -,4 $

$ \Rightarrow $ $ -,23=-,3-4n+4 $

$ \Rightarrow $ $ 4n=24 $
$ \Rightarrow $ $ n=6 $

$ \therefore $ Sum $ =\frac{6}{2}(-,3-23)=6\times -13=-78 $ Alternate Method We can do it directly as, $ [-,3-7-11-15-19-23]=-78 $

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