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?

Accepted Solution

Answer:50Step-by-step explanation:Sum of all even numbers from 2 to 100:The formula we will use is Β [tex]S_n=\frac{n}{2}(a+l)[/tex]Wheren is the number of numbersa is the first terml is the last termHere,from 2 to 100, there are 100/2 = 50 terms (n=50)first term, a = 2last term, l= 100So we have:[tex]S_n=\frac{n}{2}(a+l)\\S_{50}=\frac{50}{2}(2+100)\\=2550[/tex]Sum of all odd numbers from 1 to 99:Here, we will use a different formula for S_n.[tex]S_n=\frac{n}{2}(2a+(n-1)d)[/tex]From 1 to 99, there are 50 odd numbers (n = 50)a is the first term, a = 1d is the common difference, the difference in successive terms, the sequence is basically 1, 3, 5... so d = 3 - 1 =2Now, we substitute and find:[tex]S_{50}=\frac{50}{2}(2(1)+(50-1)(2))\\S_{50}=2500[/tex]So, subtracting Alice's result (2500) FROM Tom's (2550), we get:2550 - 2500 = 50Joe's result is 50