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Q.

What is the value of 1−2+3−4+5−...+101 are ____.

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Detailed Solution

1−2+3−4+5−...+101
⇒(1+3+5+...+101)+(−2−4−6−...−100)
⇒(1+3+5+...+101)−(2+4+6+...+100)
Now, if we assume that the given series is S, we will have:
S=1−2+3−4+5−...+101S=1−2+3−4+5−...+101
Now suppose that the series 1+3+5+...+101 is S1 and 2+4+6+...+100 is S2.
So we will have the value of S as:
S=S1−S2 …..(i)
Now we evaluate S1 .
We have:
S1=1+3+5+...+101
Now we see that this series is AP with the first term as 1, the common difference is 2 and the last term is 101.
Now in AP we know that:

a_{n}

=

a

+( n-1)d
Where a is the first term, an is the last term, d is the common difference, and n is the number of terms.
Therefore, the number of terms for S1 is:
an=a + (n-1)d
⇒101=1+ (n−1)2
⇒100=(n−1)2
⇒n=50+1
⇒n=51
Now we know that the sum of 'n' terms of AP is given as:
Sum=

\frac{n}{2} (

a+

a_{n}

)
Thus, we get the value of S1 as:

S_{1}

=

\frac{51}{2}

(1+101)
⇒

S_{1}

=

\frac{51}{2}

(102)
⇒

S_{1}

=51(51)
⇒

S_{1}

=2601
Hence, the value of

S_{1}

is 2601.
Now, we will find the value of

S_{2}

.
We have:

S_{2}

=2+4+6+...+100
Now, the number of terms for S2 are given as:
100=2+(n−1)2
⇒98=(n−1)2
⇒n=49+1
⇒n=50
Thus, the value of S2 is given as:

S_{2}

=

\frac{50}{2}

(2+100)
⇒

S_{2}

=502(102)
⇒

S_{2}

=50(51)
⇒

S_{2}

=2550
Hence, the value of

S_{2}

is 2550.
Now, putting the value of

S_{1}

and

S_{2}

in equation (i), we get:
S=

S_{1}

−

S_{2}

⇒S=2601−2550
∴ S=51
Thus, the value of S is 51.
Hence, option (1) is the first option.

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