problem: The sum of the first 5 terms of an arithmetic series is $70$.  The sum of the first 10 terms of this  arithmetic series is $210$.  What is the first term of the series?
solution: Let the first term be $a$ and the common difference be $d$.  The sum of an arithmetic series is equal to the average of the first and last term, multiplied by the number of terms.  The fifth term is $a + 4d$, so the sum of the first five terms is \[\frac{a + (a + 4d)}{2} \cdot 5 = 5a + 10d = 70,\]which implies that $a + 2d = 14$, so $2d = 14 - a$.
  
  The tenth term is $a + 9d$, so the sum of the first ten terms is \[\frac{a + (a + 9d)}{2} \cdot 10 = 10a + 45d = 210,\]which implies that $2a + 9d = 42$, so $9d = 42 - 2a$.
  
  From the equation $2d = 14 - a$, $18d = 126 - 9a$, and from the equation $9d = 42 - 2a$, $18d = 84 - 4a$, so \[126 - 9a = 84 - 4a.\]Then $5a = 42$, so $a = \boxed{\frac{42}{5}}$.