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

Answer:

50

Step-by-step explanation:

Sum of all even numbers from 2 to 100:

The formula we will use is  $S_n=\frac{n}{2}(a+l)$

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:

$S_n=\frac{n}{2}(a+l)\\S_{50}=\frac{50}{2}(2+100)\\=2550$

Sum of all odd numbers from 1 to 99:

Here, we will use a different formula for S_n.

$S_n=\frac{n}{2}(2a+(n-1)d)$

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:

$S_{50}=\frac{50}{2}(2(1)+(50-1)(2))\\S_{50}=2500$

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

2550 - 2500 = 50

Joe's result is 50