CTET 2 Math and Science Sep 2015 Paper

Question : 53 of 120

Marks: +1, -0

The additive inverse of S, where S = l – 2 + 3 – 4 + 5 – 6 + 7 – 8 + ........... + 49 – 50, is :

Solution:

{S}=1-2+3-4+5-6+7-8+. . .. . .+49-50

{S}=(1+3+5+7+. . ..49)-(2+4+6+8+. . ..+50)

=(1+3+5+7+. . .. . .,49)-2(1+2+3+4+. . ..25)

1st part 2nd part
1st part

=1+3+5+7+. . .. . ..49

n=25,a=1,d=2

{S}_{1}=\;\;{n}/{2}[2a+(n-1)d]

=\;\;{25}/{2}[2×1+(25-1)×2]

=\;\;{25}/{2}[2+48]

=\;\;{25}/{2}×50=25×25=625

2nd part

=2(1+2+3+4+5\text "________"25)

Sum of all natural number

=\;\;{n(n+1)}/{2}

n=25

{S}_{2}=\;\;{2×25(25+1)}/{2}

{S}_{2}=50×13=650

Now,

{S}={S}_{2}-{S}_{1}=650-625=25

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