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UPSC NDA 2 2019 Mathematics Solved Paper

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Question : 2 of 120

Marks: +1, -0

What is the value of

1-2+3-4+5-\dots+101\ ?

51
55
110
111

Solution: 👈: Video Solution

Concept:

If

a_1, a_2, \dots, a_n

be an AP then

S_n = a_1 + a_2 + \dots + a_n = \frac{n}{2} \times (2a + (n-1)d)

where a is the

1^{\;\text{st }\;}

term and

d

is the common difference.

If

a_1, a_2, \dots, a_n

be an AP then the general term is given by:

a_n = a + (n-1) \times d

where

a

is the

1^{\;\text{st }\;}

term and

d

is the common difference.

Calculation:

Here, we have to find the value of

1-2+3-4+5-

+101

\Rightarrow 1-2+3-4+5-

+101 = (1+3+\dots+101) - (2+4+\dots+100)

As, we can see that

(1,3,\dots,101)

is an AP with

a = 1

and

d = 2

.

\Rightarrow a_n = 101 = 1 + (n-1) \times 2

\Rightarrow n = 51

\Rightarrow S_{51} = 1+3+\dots+101 = \frac{51}{2} \times (2+50 \times 2) = 2601

Similarly,

(2,4,\dots,100)

is an AP with

a = 2

and

d = 2

\Rightarrow a_n = 100 = 2 + (n-1) \times 2

\Rightarrow n = 50

\Rightarrow S_{50} = 2+4+\dots+100 = \frac{50}{2} \times (4+49 \times 2) = 2550

\Rightarrow 1-2+3-4+5-\dots+101 = (1+3+\dots+101) - (2+4+\dots+100) = 2601-2550 = 51

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