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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 10000n+5(n1)\frac{10000}{n}+5(n-1). Given that the nth term is less than 500, show that n2101n+2000<0n^{2}-101 n+2000<0 and hence find the largest possible value of n.

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We're given an arithmetic progression with u1=au_1=a and d=10d=10. We need to express aa in terms of nn if S=10,000S=10,000 for the first nn terms of the progression. We need to prove also that

un=10000n+5(n1)u_n=\dfrac{10000}{n}+5(n-1)

For un<500u_n<500, we need to prove that n2101n+2000<0n^2-101n+2000<0 and determine its largest possible value of nn.

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