Question

39.
Find the value of expression given below
1 - 2 + 3-4+5-6 + ... + 97–98 + 99 - 100
+ 101
(2) 51
(1) 50
(3) 200
(4) Can't find​

Answer 1

Answer:

Step-by-step explanation: question can be written as

1 +( - 2 + 3) + (-4+5) (-6 +7) ... + 97+  (-98 + 99) + (- 100 + 101)

= 1 + 1×(number of pairs)           {number of pairs = number of even numbers}

=1+ 1×(50)

= 51

Answer 2

Answer:

Option (2)

Step-by-step explanation:

Given :-

1 -2 + 3-4+5-6 + ... + 97–98 + 99 - 100 + 101

To find :-

Find the value of the expression ?

Solution :-

Given that :

1-2+3-4+5-6+...+97-98+99-100+101

It can be written as

(1+3+5+...+99+101)+(-2-4-6-...-98-100)

(1+3+5+..+99+101) - (2+4+6+...+98+100)

Now , On taking 1+3+5+...+99+101

First term (a) = 1

Common difference = 3-1 = 2

Since the common difference is same throughout the series they are in the AP

Last term = (l) = 101

We know that

nth term of an AP = an = a+(n-1)d

Let an = 101

=> a+(n-1)d = 101

=> 1+(n-1)(2) = 101

=> 1+2n-2 = 101

=>2n -1 = 101

=>2n = 101+1

=>2n = 102

=> n = 102/2

=> n = 51

Number of terms = 51

We know that

Sum of first n terms = Sn = (n/2)[2a+(n-1)d]

=> S 51 = (51/2)[2(1)+(51-1)(2)]

=> S 51 = (51/2)[2+50(2)]

=> S 51 = (51/2)[2+100]

=> S 51 = (51/2)(102)

=> S 51 = 51×51

=> S 51 = 2601 -------------------(1)

On taking

2+4+6+...+98+100

First term (a) = 2

Common difference = 4-2 = 2

Since the common difference is same throughout the series they are in the AP

Last term = (l) = 100

We know that

nth term of an AP = an = a+(n-1)d

Let an = 100

=> a+(n-1)d = 100

=> 2+(n-1)(2) = 100

=> 2+2n-2 = 100

=>2n = 100

=> n = 100/2

=> n = 50

Number of terms = 50

We know that

Sum of first n terms = Sn = (n/2)[2a+(n-1)d]

=> S 50 = (50/2)[2(2)+(50-1)(2)]

=> S 50 = (50/2)[4+49(2)]

=> S 50 = (50/2)[4+98]

=> S 50 = (50/2)(102)

=> S 50 = 50×51

=> S 50 = 2550 -------------------(2)

Now we have

(1+3+5+..+99+101) - (2+4+6+...+98+100)

From (1)&(2)

=> 2601 - 2550

=> 51

Answer:-

The value of 1-2+3-4+5-6+...+97-98+99-100+101 is 51

Used formulae:-

• nth term of an AP = an = a+(n-1)d

• Sum of first n terms = Sn =
• (n/2)[2a+(n-1)d]

• Sn = (n/2)[a+an]

• a = First term

• d = Common difference

• n = number of terms

• an = nth term or general term or last term

• Sum of first n even numbers = n(n+1)

• Sum of first n odd numbers = n^2