An arithmetic progression has first term a and common difference 10. The sum of the first n terms of the progression is 10,000. Express a in terms of n, and show that the nth term of the progression is:
10,000/n + 5(n-1)
Given that the nth term is less than 500, show that n^2 - 101n + 2000 < 0 and hence find the largest possible value of n.
Let d = 10
I tried it using the formula of sum:
S = n/2 (2a + (n-1)d)
10,000 = n/2 (2a + 10n - 10)
10,000 = na + 5n^2 - 5n
10,000 - 5n^2 + 5n = na
=> a = 10,000/n + 5(1 - n)
The nth term should be: a + (n-1)d => 10,000/n + 5(1-n) + 10n - 10
= 10,000/n - 5 + 5n => 10,000/n + 5(n - 1) (shown)
Now ... I don't know how to solve the rest of it.