Sum of geometric progression

Question

Given a geometric sequence whose sum of the first ten terms is 4, and whose sum from the 11th to the 30th term is 48. Find the sum the 31st to the 60th term.

Answer 1

We may write the two conditions as $$\begin{align} \frac{a}{r-1}(r^{10}-1) &=4,\;\; \text{and}\\ \frac{a r^{10}}{r-1}(r^{20}-1) &=48. \end{align} $$ Dividing the second equation by the first yields$$ r^{10}(r^{10}+1) = 12. $$ We let $x = r^{10}$ and solve $x(x+1)=12$ to find that $r^{10} = 3$ or $-4$. Geometric progressions are usually defined as real sequences, so we will assume $r^{10} = 3$. One could proceed with the $-4$ case in the same way.

Substituting $r^{10} = 3$ into the first equation above yields$$ \frac{a}{r-1} = 2.$$ Therefore the sum of the $31^\mathrm{th}$ through $60^\mathrm{th}$ terms is $$ \frac{a r^{30}}{r-1}(r^{30}-1) = 2 \cdot 3^3 \cdot (3^3 - 1) = 1404.$$