In an arithmetic progression the sum of the first 10 terms is 155 and the sum of the first 20 terms is 610. The sum of its first 30 terms is:

Using Sₙ = $(n/2)(2a+(n-1)d):5(2a+9d)=155$ gives $2a+9d=31$, and $10(2a+19d)=610$ gives $2a+19d=61$. Subtracting, $10d=30$, so $d=3$ and $2a=4$, i.e. $a=2$. Then S₃₀ = $15(4+29\times 3)=15\times 91=1365$. (1065 treats S₁₀, S₂₀, S₃₀ as themselves being in arithmetic progression, giving 610 + 455; 765 adds the two given sums; 1220 doubles S₂₀.) The progression is 2, 5, 8, … and S₁₀ = $5(4+27)=155$.