Lakshman Bhaiya
asked
in Quantitative Aptitude
Apr 3, 2020
recategorized
Nov 8, 2020
by Krithiga2101

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$1-6+2-7+3-8+\cdots\ to\ 100\ terms$

$=\underbrace{1-6}_{-5}+\underbrace{2-7}_{-5}+\underbrace{3-8}_{-5}+\cdots +\underbrace{50-55}_{-5}$

$= (1+2+\cdots + 50) - (6+7+\cdots +55)$

$= (1+2+3+4+5)+\underbrace{(6+7+\cdots +50) - (6+7+\cdots +50)}_0-(51+52+53+54+55)$

$= (1+2+3+4+5)-(51+52+53+54+55) $

$=\cfrac{5}{2}\Bigl[2(1)+(5-1)(1)\Bigr] - \cfrac{5}{2}\Bigl[2(51)+(5-1)(1)\Bigr]$

$\left( \because Sum\ of\ n\ terms\ of\ A.P.\ sequence\ is,\ \cfrac{n}{2}\Bigl[2a+(n-1)d\Bigr]\right)$

$=\cfrac{5}{2}(6) - \cfrac{5}{2}(106)$

$=5(3)-5(53)$

$=15-265$

$=-250$

$=\underbrace{1-6}_{-5}+\underbrace{2-7}_{-5}+\underbrace{3-8}_{-5}+\cdots +\underbrace{50-55}_{-5}$

$= (1+2+\cdots + 50) - (6+7+\cdots +55)$

$= (1+2+3+4+5)+\underbrace{(6+7+\cdots +50) - (6+7+\cdots +50)}_0-(51+52+53+54+55)$

$= (1+2+3+4+5)-(51+52+53+54+55) $

$=\cfrac{5}{2}\Bigl[2(1)+(5-1)(1)\Bigr] - \cfrac{5}{2}\Bigl[2(51)+(5-1)(1)\Bigr]$

$\left( \because Sum\ of\ n\ terms\ of\ A.P.\ sequence\ is,\ \cfrac{n}{2}\Bigl[2a+(n-1)d\Bigr]\right)$

$=\cfrac{5}{2}(6) - \cfrac{5}{2}(106)$

$=5(3)-5(53)$

$=15-265$

$=-250$