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..

  • 1
    $\begingroup$ Going forward I might suggest - mathjax $\endgroup$ – Antonio Hernandez Maquivar Sep 9 '16 at 19:57
  • 1
    $\begingroup$ 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

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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.