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To find the common difference d we note that the $\\var{n1}$th term is $\\var{t1}$ and the $\\var{n2}$th term is $\\var{t2}$ so we can find the common difference d by noting $\\var{t2}-\\var{t1}=(\\var{n2}-\\var{n1})\\times d$, hence common difference $= \\var{d}$

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To find the least positive value for n for which the sum of the first n terms of the series exceeds 1000, we recall that the sum of an arithmetic series can be found with the formula $S_n = \\frac{n(2a_1 + (n-1)\\times d)}{2}$ where $a_1$ denotes the first term and d denotes the common difference. Now we want to find the smallest positive $n$ such that $S_n > 1000$. This is now a matter of solving a quadratic, which is left as an exercise. 

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What is the common difference?

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Determine the least positive value for $n$ for which the sum of the first $n$ terms of the series exceeds 1000.

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In an arithmetic series, the $\\var{n1}$th term is $\\var{t1}$ and the $\\var{n2}$th term is $\\var{t2}$

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