If the sum of the first $10$ terms of an arithmetic progression is $155$ and the sum of the first $20$ terms is $610$, then the sum of the first $30$ terms is:

Solution

1. The sum of the first $n$ terms is given by $S_n=\frac{n}{2}(2a+(n-1)d)$.
2. For $n=10$, $$\begin{array}{rl}{S}_{10}& =5(2a+9d)\\ & =155\end{array}$$, which simplifies to $2a + 9d = 31$.
3. For $n=20$, $$\begin{array}{rl}{S}_{20}& =10(2a+19d)\\ & =610\end{array}$$, which simplifies to $2a + 19d = 61$.
4. Subtracting the first equation from the second gives $10d = 30$, so $d=3$, and substituting back gives $a=2$.
5. The sum of the first 30 terms is $$\begin{array}{rl}{S}_{30}& =\frac{30}{2}(2(2)+29(3))\\ & =15(4+87)\\ & =15(91)\\ & =1365\end{array}$$.
6. Hence the answer is (A).