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**struck** 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.

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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.