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Question

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

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Answered 2 years ago

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1 of 10We're given an arithmetic progression with $u_1=a$ and $d=10$. We need to express $a$ in terms of $n$ if $S=10,000$ for the first $n$ terms of the progression. We need to prove also that

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

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

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