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Consider the arithmetic series $-6,1,8,15....$ Find the least number of terms so that the sum of the series is greater than $1000$.

I don't know how to do it,the only thing I got is this:

$a=-6 \\ d=7$

$n^{\text{th}}$ term is given by $a+(n-1)*d =-6+7n-7 =7n-13$

Please help..

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    $\begingroup$ Going forward I might suggest - mathjax $\endgroup$ – Antonio Hernandez Maquivar Sep 9 '16 at 19:57
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    $\begingroup$ There is a formula to get the sum of terms of AP. [n*(2*a+(n-1)*d)]/2. Basically it means n*(average of first and nth term). Try with this formula $\endgroup$ – user3219492 Sep 9 '16 at 19:59
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For your $a_k=7k-13$, consider that $\sum_{k=1}^na_k=\frac{1}{2}n(a_1+a_n)$. You want this to be greater than $1000$, hence \begin{equation} \frac{1}{2}n(a_1+a_n)=\frac{1}{2}n(-6+7n-13)>1000 \end{equation}

Can you take it from here?

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