In an arithmetic series, if $a_{4}\; +\; a_{7}\; =\; 30$ what is the sum of the first 100 terms

If you are given an arithmetic series whose 4th term and 7th term add up to 30, how would you find the sum of the first 100 terms?

-
start with $a_1$. Since it's an arithmetic series, $a_2=a_1+d$ for some $d$, $a_3=a_2+d=a_1+2d$, and thus $a_4=a_1+3d$. Then $a_7=a_1+6d$. – Eleven-Eleven Dec 11 '13 at 0:10
Yes, I got those and I also know that $s_{100}\; =\; 50\left( a_{1}\; +\; a_{100} \right)$, but I am unsure how to proceed from there. – 1110101001 Dec 11 '13 at 0:12
Rewrite $a_n=a+(n-1)d$. Therefore, $2a+9d=30$. $S_{100}=\frac{99}{2}(2a+99d)$. Hmm, I'm stuck. Are you sure its not the first 10 terms? – Sachin_ruk Dec 11 '13 at 0:12
There simply isn't enough information to determine the sum of the first $100$ terms. – Daniel Fischer Dec 11 '13 at 0:13
either $a_1$ or $d$ must be given or else the solution will depend on at least one of them... – Sergio Parreiras Dec 11 '13 at 0:14

As $a_4=a+3d$ and $a_7=a+6d$, we are given that $2a+9d=30$. The sum of the first $100$ terms is $100a+\frac {99\cdot 100}2d=100a+4950d$. We are close. If it were $495d$, we could say the last sum was $50(2a+9d)=1500$. I suspect the problem poser dropped the last zero.