AP and GP crossover

When a sequence satisfies both the definition of an arithmetic progression and a geometric progression, we can use their defining properties to find all such sequences. This problem, from the HSC Advanced Mathematics sequences topic, challenges your ability to merge two fundamental models of sequences and see the intersection: the constant sequence. By setting up the common difference $d$ and common ratio $r$, you'll gain insight into why only the constant sequence can be both an AP and a GP.

Problem Statement

A sequence is both arithmetic and geometric. Given $a_1=6$ and $a_3=6$, find all possible sequences.

---

Hints

Let AP difference be $d$ and GP ratio be $r$. Use both models on $a_3$.

---

Solutions

We start by modelling the sequence as an arithmetic progression. Let $d$ be the common difference.
AP gives

$$a_3=a_1+2d=6+2d=6 \implies d=0,$$

so AP is constant: $a_n=6$.

Now for the geometric progression. Let $r$ be the common ratio.
For GP,

$$a_3=a_1r^2=6r^2=6 \implies r^2=1 \implies r=\pm 1.$$

The case $r=1$ yields the constant $6$ sequence, which already satisfies the arithmetic condition. But we must check $r=-1$ carefully.
If $r=-1$, terms alternate $6,-6,6,\ldots$ and are not arithmetic.

Hence the only sequence that is both AP and GP is

$$a_n=6\quad \text{for all }n,$$

with $d=0$ and $r=1$.

---

Takeaways

• Only constant sequences (or the zero sequence, excluded here) can be simultaneously arithmetic and geometric – a powerful structural fact.
• Solving the two definitions together forces $d=0$ and $r=1$; the squared equation $r^2=1$ introduces an extraneous candidate $r=-1$, which must be discarded because it breaks the arithmetic condition.
• Always test each solution against all original constraints – geometry can hide arithmetic incompatibilities, and vice‑versa.

---

Further Readings

HSC Inequalities, HSC Sequences, HSC Functions, HSC Probability