jmc

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 $a=\boxed{\frac{42}{5}}$.