Question

14. The sum of all the whole numbers from 1 to 30 is subtracted from the sum of
all the whole numbers from 31 to 60 as shown.
(31 + 32 +33 +...+60) - (1 +2+3+...+ 30)
What will be the result?
A. 1200
B. 900
C. 120
D. 30

Answer 1

Given :  The sum of all the whole numbers from 1 to 30 is subtracted from the sum of all the whole numbers from 31 to 60

To Find : What will be the result?

A. 1200

B. 900

C. 120

D. 30

Solution:

Simple method :

(31 + 32 +33 +...+60) - (1 +2+3+...+ 30)

= (31 - 1) + (32 - 2) + (33 - 3) + .. + .. + ...  + (60 - 30)

= 30 + 30 + 30  + .. + .. + ...  + 30

= 30 * 30

= 900

Detailed method :

31 + 32 +33 +...+60

AP a = 31 d = 1  n = 30  L = 60

Sum = (30/2)(31 + 60)   = 15(91)

1 +  2 + 3 +...+30

AP a =  1 d = 1  n = 30 L = 30

Sum = (30/2)( 1 + 30)   = 15(31)

15(91) - 15(31)

= 15(91 - 31)

= 15(60)

= 900

Learn More:

In an A.P if sum of its first n terms is 3n square +5n and it's Kth term ...

brainly.in/question/8236011

Find the sum of frist 51 terms of the AP whose 2nd term is 2 and 4th ...

brainly.in/question/7655866