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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?

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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
up vote 1 down vote accepted

The way you would approach this would be to find the first term, find the delimiter, and then find the last term. The sum will be (50)(a1+a100). It's not really possible to determine what a1/a100 are, nor the delimiter because there isn't enough information.

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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.

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