Question

Tom added all even numbers from 2 to 100. Alice's added all odd numbers from 1 to 99. Then joe subtracted Alices result from Tom's result. What is joe's result?

Answer 1

50

Explanation:

Sum of all even numbers from 2 to 100:

The formula we will use is
`S_n=(n)/(2)(a+l)`

Where

n is the number of numbers

a is the first term

l is the last term

Here,

from 2 to 100, there are 100/2 = 50 terms (n=50)

first term, a = 2

last term, l= 100

So we have:

`S_n=(n)/(2)(a+l)\\S_(50)=(50)/(2)(2+100)\\=2550`

Sum of all odd numbers from 1 to 99:

Here, we will use a different formula for S_n.

`S_n=(n)/(2)(2a+(n-1)d)`

From 1 to 99, there are 50 odd numbers (n = 50)

a is the first term, a = 1

d is the common difference, the difference in successive terms, the sequence is basically 1, 3, 5... so d = 3 - 1 =2

Now, we substitute and find:

`S_(50)=(50)/(2)(2(1)+(50-1)(2))\\S_(50)=2500`

So, subtracting Alice's result (2500) FROM Tom's (2550), we get:

2550 - 2500 = 50

Joe's result is 50