If "a" is the first term of an AP and "d" is common difference, then sum of first "n" terms is "an+\\frac{n(n-1)}{2}d".
Since sum of first five terms is "35" and sum of next five terms is "85", we obtain that sum of first ten terms is "120".
So "35=5a+\\frac{5(5-1)}{2}d=5a+10d" and "120=10a+\\frac{10(10-1)}{2}d=10a+45d".
We have "120-35\\cdot 2=(10a+45d)-2(5a+10d)", that is "50=25d", so "d=2".
Then we have "35=5a+10d=5a+10\\cdot 2", so "a=3"
Answer: the first term is "3", the common difference is "2"
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