An algebra problem by Ajala Singh

finding sum of n number=n(n+1)/2. Here n is 100. So, 100*101/2=5050. We have to find avg for 100 postive intergers. so 5050/100=50.5

first + last integer divided by 2.

1+100=2+99=3+98=4+97=...=49+52=50+51=101. There're 50 addition (from 1+100 to 50+51) that equals to 50 times 101. For the first 100 positive integers we can get (50 x 101)/100 = 0.5 x 101=50.5.

$s=\dfrac{n}{2}\left(a_1+a_n\right)=\dfrac{100}{2}\left(1+100\right)=5050$

$average=\dfrac{5050}{100}=$ $\color{#D61F06}\boxed{50.5}$

Ignoring the terms between 1 and 100 since for example.

(1+100 )/2 = 50.5

Because

(2+99)/2 = 50.5 , (3+98)/2 = 50.5 , . . . . , (50+51)/2 = 50.5

50(1+100)/100=50(101)/100=5050/100=50.5

using formula for sum of all consecutive integers n(n+l)/2 using n = 100 then the sum = 5050 solving for arithmetic mean = sum of all quantities/number of quantitie = 5050/100 A.M. = 50.5

n(n+1)/2 100 100(100+1) /2 100