A Geometric Progression (GP) is built by multiplying, not adding. Each term is the previous one times a fixed number called the common ratio $r$. So a GP reads $a,\ ar,\ ar^2,\ ar^3, \dots$, where $a$ is the first term and $r \ne 0$. The test for a GP is that the ratio $\dfrac{a_n}{a_{n-1}}$ is the same for every $n$.
The general ($n$th) term records how many times you have multiplied by $r$:
$$a_n = ar^{\,n-1}$$
The sum of the first $n$ terms has two forms; pick the one that keeps the numerator positive:
$$S_n = \dfrac{a\left(r^{n} - 1\right)}{r - 1}\ \ (r > 1) \qquad S_n = \dfrac{a\left(1 - r^{n}\right)}{1 - r}\ \ (r < 1)$$
When $r = 1$ every term equals $a$, so $S_n = na$ — do not use the formula then, since it divides by zero.
Something remarkable happens when the ratio is a proper fraction. If $|r| < 1$, the terms shrink toward zero and the running total settles on a finite limit even though the series never stops. This is the sum to infinity:
$$S_\infty = \dfrac{a}{1 - r}, \qquad |r| < 1$$
The Geometric Mean (GM) of two positive numbers $a$ and $b$ is the value $G$ that makes $a, G, b$ a GP. Equating ratios ($\tfrac{G}{a} = \tfrac{b}{G}$) gives $G^2 = ab$, so:
$$G = \sqrt{ab}$$
To insert $k$ geometric means between $a$ and $b$ is to find $k$ numbers making the whole list a GP of $k+2$ terms. Since $b$ is then the $(k+2)$th term, $b = ar^{k+1}$, giving the ratio $r = \left(\dfrac{b}{a}\right)^{1/(k+1)}$, and the $i$th inserted mean is $G_i = a\,r^{\,i}$. A pretty fact mirroring the AP case: the product of all $k$ inserted means equals $G^k$, the $k$th power of the single GM of $a$ and $b$.
There is a clean and famous relationship between the two means of the same pair of positive numbers — the AM–GM inequality:
$$\text{AM} \ge \text{GM} \quad\Longleftrightarrow\quad \dfrac{a + b}{2} \ge \sqrt{ab}$$
Equality holds only when $a = b$. The proof is one line: $\left(\sqrt{a} - \sqrt{b}\right)^2 \ge 0$ expands to $a + b \ge 2\sqrt{ab}$. Two further identities link the means: $A \cdot H = G^2$ (with the harmonic mean $H$) and, for two numbers, $G = \sqrt{A H}$ — so the GM is itself the geometric mean of the AM and HM.
Deeper Insight — multiplication, growth, and the surprise of a finite infinite sum: Where the AP is the discrete cousin of a straight line, the GP is the discrete cousin of exponential growth and decay. A constant ratio means each term is a fixed percentage larger (or smaller) than the last, which is exactly how compound interest, population growth, radioactive decay, and bouncing balls behave in the real world. The deepest idea in this topic is the sum to infinity. It feels paradoxical that adding infinitely many positive numbers can give a finite answer, but the resolution is that when $|r| < 1$ the terms vanish so fast that the leftover "tail" you keep adding becomes negligibly small — the partial sums creep ever closer to $\tfrac{a}{1-r}$ without ever overshooting it. This single mechanism is what lets a recurring decimal like $0.\overline{3}$ equal the exact fraction $\tfrac{1}{3}$. Finally, the AM–GM inequality is one of the most powerful tools in all of mathematics for proving things are as large or as small as possible; remembering that the two means coincide only when the numbers are equal is the key that unlocks countless optimisation arguments.
Find the 8th term of the GP $5, 10, 20, 40, \dots$
Solution- Identify $a = 5$ and $r = \dfrac{10}{5} = 2$.
- Apply $a_n = ar^{n-1}$ with $n = 8$.
- $a_8 = 5 \times 2^{7} = 5 \times 128 = 640$.
Answer: $a_8 = 640$.
Find the sum of the first $6$ terms of the GP $2, 6, 18, \dots$
Solution- Here $a = 2$, $r = 3$ (so use the $r > 1$ form).
- $S_n = \dfrac{a(r^n - 1)}{r - 1} = \dfrac{2(3^6 - 1)}{3 - 1}$.
- $3^6 = 729$, so $S_6 = \dfrac{2(729 - 1)}{2} = 729 - 1 = 728$.
Answer: $S_6 = 728$.
Find the sum to infinity of $9 + 3 + 1 + \dfrac{1}{3} + \dots$
Solution- First term $a = 9$ and ratio $r = \dfrac{3}{9} = \dfrac{1}{3}$, with $|r| < 1$ so the sum exists.
- $S_\infty = \dfrac{a}{1 - r} = \dfrac{9}{1 - \tfrac{1}{3}} = \dfrac{9}{\tfrac{2}{3}}$.
- $= 9 \times \dfrac{3}{2} = \dfrac{27}{2} = 13.5$.
Answer: $S_\infty = \dfrac{27}{2}$.
Express the recurring decimal $0.\overline{4} = 0.4444\dots$ as a fraction using a GP.
Solution- Write it as the series $\dfrac{4}{10} + \dfrac{4}{100} + \dfrac{4}{1000} + \dots$
- This is a GP with $a = \dfrac{4}{10}$ and $r = \dfrac{1}{10}$, and $|r| < 1$.
- $S_\infty = \dfrac{a}{1 - r} = \dfrac{4/10}{1 - 1/10} = \dfrac{4/10}{9/10} = \dfrac{4}{9}$.
Answer: $0.\overline{4} = \dfrac{4}{9}$.
Find the geometric mean of $4$ and $25$.
Solution- The GM of two positive numbers is $G = \sqrt{ab}$.
- $G = \sqrt{4 \times 25} = \sqrt{100}$.
- $= 10$.
Answer: $G = 10$.
The AM of two positive numbers is $10$ and their GM is $8$. Find the numbers.
Solution- $\dfrac{a + b}{2} = 10 \Rightarrow a + b = 20$, and $\sqrt{ab} = 8 \Rightarrow ab = 64$.
- The numbers are roots of $t^2 - (a+b)t + ab = 0$, i.e. $t^2 - 20t + 64 = 0$.
- Factorise: $(t - 4)(t - 16) = 0 \Rightarrow t = 4$ or $t = 16$.
- Check: $\text{AM} = 10 \ge \text{GM} = 8$, consistent with AM–GM.
Answer: The numbers are $4$ and $16$.
The 3rd term of a GP is $24$ and the 6th term is $192$. Find the first term and the common ratio.
Solution- $a_3 = ar^2 = 24$ and $a_6 = ar^5 = 192$.
- Divide: $\dfrac{ar^5}{ar^2} = \dfrac{192}{24} \Rightarrow r^3 = 8 \Rightarrow r = 2$.
- Then $ar^2 = 24 \Rightarrow 4a = 24 \Rightarrow a = 6$.
Answer: $a = 6$ and $r = 2$.
Insert three geometric means between $2$ and $32$.
Solution- Inserting $3$ means makes a $5$-term GP, so $32$ is the 5th term: $32 = 2\,r^{4}$.
- $r^4 = 16 \Rightarrow r = 2$ (taking the positive ratio).
- The means are $2\times 2 = 4$, $4\times 2 = 8$, $8\times 2 = 16$.
Answer: The three geometric means are $4, 8, 16$.
A ball is dropped from a height of $12$ m and on each bounce rises to $\dfrac{2}{3}$ of the previous height. Find the total vertical distance it travels before coming to rest.
Solution- The first drop is $12$ m. After that each up-and-down adds twice a bounce height.
- Bounce heights form a GP $12\cdot\tfrac{2}{3}, 12\cdot(\tfrac{2}{3})^2, \dots = 8, \tfrac{16}{3}, \dots$ with $a = 8$, $r = \tfrac{2}{3}$.
- Total upward + downward after the first drop $= 2\,S_\infty = 2\cdot\dfrac{8}{1 - \tfrac{2}{3}} = 2\cdot\dfrac{8}{1/3} = 2\times 24 = 48$.
- Add the initial drop: $12 + 48 = 60$.
Answer: The ball travels a total of $60$ m.
Find three numbers in GP whose sum is $21$ and the sum of whose squares is $189$.
Solution- Take the symmetric form $\dfrac{a}{r},\ a,\ ar$.
- Sum: $\dfrac{a}{r} + a + ar = 21 \Rightarrow a\left(\tfrac{1}{r} + 1 + r\right) = 21$.
- Sum of squares: $\dfrac{a^2}{r^2} + a^2 + a^2 r^2 = 189$.
- Try $a = 6$: then $\tfrac{1}{r} + 1 + r = \tfrac{21}{6} = \tfrac{7}{2} \Rightarrow r + \tfrac{1}{r} = \tfrac{5}{2} \Rightarrow 2r^2 - 5r + 2 = 0 \Rightarrow r = 2$ or $\tfrac{1}{2}$.
- With $a = 6$, $r = 2$: the numbers are $3, 6, 12$ (squares $9 + 36 + 144 = 189$, checks out).
Answer: The numbers are $3, 6, 12$.
If $a, b, c$ are in GP, prove that $b^2 = ac$, and verify for $3, 6, 12$.
Solution- In a GP the ratio between consecutive terms is constant: $\dfrac{b}{a} = \dfrac{c}{b}$.
- Cross-multiplying gives $b^2 = ac$ — the condition for three numbers to be in GP.
- Verify with $3, 6, 12$: $b^2 = 6^2 = 36$ and $ac = 3 \times 12 = 36$. Equal.
Answer: $b$ is the geometric mean, so $b^2 = ac$; verified since $36 = 36$.
Evaluate the sum to infinity $\dfrac{2}{3} - \dfrac{4}{9} + \dfrac{8}{27} - \dots$
Solution- First term $a = \dfrac{2}{3}$; ratio $r = \dfrac{-4/9}{2/3} = -\dfrac{2}{3}$, and $|r| = \tfrac{2}{3} < 1$.
- $S_\infty = \dfrac{a}{1 - r} = \dfrac{2/3}{1 - (-2/3)} = \dfrac{2/3}{5/3}$.
- $= \dfrac{2}{3} \times \dfrac{3}{5} = \dfrac{2}{5}$.
Answer: $S_\infty = \dfrac{2}{5}$.