Question

There are 20 consecutive even numbers. How much bigger is the sum of the 10 larger ones than the sum of the ten smaller ones?

Answer 1

Answer:

Step-by-step explanation:

Given There are 20 consecutive even numbers. How much bigger is the sum of the 10 larger ones than the sum of the ten smaller ones?

Let the ten smaller even numbers be n, n+2, n+4, n+6...........n+18 and let the ten larger numbers be n+20,n+22,n+24.............n+38

Subtract both the numbers we get

n+20-n = 20

n+22-n-2 = 20

So like this we get 20 or 20 x 10 = 200

Answer 2

Let us consider that x is an even number , then next 19 consecutive even numbers are : (x + 2), (x + 4) , (x + 6) , ..........., {x + (20 - 1) × 2 }

e.g., 20 consecutive numbers are : x, (x + 2), (x + 4), (x + 6), ........ (x + 38)

sum of first ten consecutive even numbers = x + (x + 2) + (x + 4) + ....... (x + 18)..........(1)


again, sum of next 10 consecutive even numbers = (x + 20) + (x + 22) + (x + 24) + ...... + (x + 38) ..........(2)


difference equation (1) by (2),

we get, 20 + (22 - 2) + (24 - 4) + (26 - 6) + .... + (38- 18)

= 20 + 20 + 20 + 20 ... 10 times

= 200

hence, sum of 10 larger ones is 200 much greater than 10 smaller ones .