Question

4. Find :
(a) 200 +(-174)+(-26)
(c) (-18)+(+25)+(-37)
(b) 4+(-99)+(-101)+96
(d) (-100)+(-99)+(-98)+...98 +99 +100​

Answer 1

Step-by-step explanation:

(a) Solution:

= 200 +(-174)+(-26)

= 200-174-26

= 200-200

= 0

(C) Solution:

= (-18)+(+25)+(-37)

= -18 +25 -37

= -55+25

= -30

(b) Solution:

= 4+(-99)+(-101)+96

= 4-99-101+96

= 100-200

= -100

(d) Solution

The given series of number are (-100)+(-99)+(-98)+...98 +99 +100

These numbers form an AP ,where

First term, a = (-100). ;

Common difference,d = {(-99)-(-100)} = ( -99+100)= 1 ;

Last term ,l = 100

So, Tn = a+(n-1)d

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

=> 100 +100 = n-1

=> 200+1 = n

=> n= 201

So, Sn = n/2(a+l)

=> S201 = 201/2(-100+100)

=> S201 = 201/2*0

=> S201 = 0

Hence, Sum of all term is 0

Therefore, answer is 0

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