Finding the number of terms in an arithmetic series

Question

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..

Answer 1

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?