Architecture of GP sums

In HSC Mathematics Advanced and Extension 1, sequences and series questions often test your ability to manipulate the sum formula and factorise expressions. This problem explores a neat pattern in geometric series — the ratio of sums — and shows how algebraic factorisation can unlock seemingly complex relationships. By working through it, you’ll gain confidence with difference‑of‑squares factorisation and see how to connect different GPs.

Problem Statement

Let a GP have first term $a\neq0$, ratio $r\neq 0,\pm1$, and partial sums $S_n$.

• Prove $\dfrac{S_{2n}}{S_n}=r^n+1$.
• Given $S_{12}=65S_6$ and second term $ar=12$, find possible $(a,r)$.
• If $\Sigma_n$ is the partial-sum sequence of a GP with first term $a$ and ratio $r^2$, show

$$\frac{\Sigma_n}{S_n}=\frac{r^n+1}{r+1}.$$

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Hints

Use $S_n=\dfrac{a(r^n-1)}{r-1}$ and difference-of-squares factorization.

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Solutions

We start with the standard formula for the sum of the first $n$ terms of a GP: $S_n = \frac{a(r^n-1)}{r-1}$. Applying this to $S_{2n}$ gives $S_{2n} = \frac{a(r^{2n}-1)}{r-1}$. Their ratio simplifies as follows:

$$\frac{S_{2n}}{S_n}=\frac{r^{2n}-1}{r^n-1}=r^n+1.$$

Notice how $r^{2n}-1$ factors via difference of squares into $(r^n-1)(r^n+1)$, allowing the denominator to cancel neatly, leaving $r^n+1$.

We are given $S_{12}=65 S_6$. Comparing this with our derived identity, we see that with $n=6$, $S_{2n}/S_n = S_{12}/S_6 = r^6+1$. So we set up

$r^6+1=65\implies r^6=64\implies r=\pm2$.

Since the second term is $ar=12$, we substitute $r=\pm 2$ to find $a$. For $r=2$, $a \cdot 2 = 12 \implies a=6$. For $r=-2$, $a \cdot (-2) = 12 \implies a=-6$.

Thus the possible pairs are $(a,r)=(6,2)$ or $(-6,-2)$.

Now consider a new GP with the same first term $a$ but ratio $r^2$. Its partial sum $\Sigma_n$ is given by the sum formula with ratio $r^2$: $\Sigma_n = \frac{a(r^{2n}-1)}{r^2-1}$. We relate this back to $S_n$ by writing

$$\Sigma_n=\frac{a(r^{2n}-1)}{r^2-1},$$

thus

$$\frac{\Sigma_n}{S_n} =\frac{r^{2n}-1}{(r+1)(r^n-1)} =\frac{r^n+1}{r+1}.$$

Again, $r^{2n}-1 = (r^n-1)(r^n+1)$ allows cancellation, giving the final neat form.

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Takeaways

• The ratio $S_{2n}/S_n = r^n+1$ emerges from difference-of-squares factorisation — a recurring trick in GP problems.
• Given relationships between partial sums, always try to express them in terms of $r^n$ rather than expanding fully.
• When a GP's ratio is squared, the sum formula simplifies neatly, and you can often connect it back to the original sum via algebraic manipulation.

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Further Readings

HSC Polynomials, HSC Sequences, HSC Proofs, HSC Combinatorics