Geometric progressions

Multiply, do not add

Fold a sheet of paper $0.1$ mm thick in half. It becomes $0.2$ mm thick. Fold again and it is $0.4$ mm, then $0.8$ mm. Each fold doubles the thickness. After $10$ folds the stack would be $0.1 \times 2^{10} = 102.4$ mm thick.

The thicknesses $0.1, 0.2, 0.4, 0.8, \ldots$ do not go up by a fixed amount. They go up by a fixed factor. Such a list is a geometric progression.

The definition

A sequence $a_1, a_2, a_3, \ldots$ of non-zero numbers is a geometric progression (GP) if each term, divided by the one before it, gives the same number:

The fixed number $r$ is the common ratio. The first term is usually called $a$.

• $3, 12, 48, 192, \ldots$ is a GP with $a = 3$ and $r = 4$.
• $81, 27, 9, 3, \ldots$ is a GP with $a = 81$ and $r = \tfrac{1}{3}$.
• $5, -10, 20, -40, \ldots$ is a GP with $a = 5$ and $r = -2$. A negative ratio flips the sign each time.

No term of a GP can be $0$, and $r$ cannot be $0$ either. Otherwise the ratio of two terms would not exist.

The nth term

Write the terms in powers of $r$:

The first term carries $r^0$, the second $r^1$, the third $r^2$. So term number $n$ carries $r^{n - 1}$:

For $3, 12, 48, \ldots$, the 6th term is $3 \times 4^5 = 3 \times 1024 = 3072$.

Set $r = 1$. Every bar is the same height: a constant sequence is a GP with ratio $1$.

Finding a GP from two terms

The 3rd term of a GP is $12$ and the 6th term is $96$. Find the 10th term.

Step 1. Write both terms. $a r^2 = 12$ and $a r^5 = 96$.

Step 2. Divide. $\dfrac{a r^5}{a r^2} = r^3 = \dfrac{96}{12} = 8$, so $r = 2$.

Step 3. Find a. $a \times 4 = 12$, so $a = 3$.

Step 4. The 10th term. $3 \times 2^9 = 3 \times 512 = 1536$.

Dividing one term by another cancels $a$. That is the quickest first step.

Three numbers in GP

When three numbers in GP have a known product, write them as $\dfrac{a}{r}$, $a$ and $ar$. The product is then $a^3$, and $r$ drops out.

Example. Three numbers in GP have product $1000$ and sum $35$. Find them.

Step 1. The product. $a^3 = 1000$, so $a = 10$.

Step 2. The sum. $\dfrac{10}{r} + 10 + 10r = 35$. Multiply by $r$: $10r^2 - 25r + 10 = 0$, or $2r^2 - 5r + 2 = 0$.

Step 3. Solve. $(2r - 1)(r - 2) = 0$, so $r = 2$ or $r = \tfrac{1}{2}$.

Answer: the numbers are $5$, $10$ and $20$, in one order or the other.

A test for three terms

Three non-zero numbers $a$, $b$, $c$ are in GP exactly when $\dfrac{b}{a} = \dfrac{c}{b}$, that is,

For example, $4, 10, 25$ is a GP, since $10^2 = 100 = 4 \times 25$.

Growth by a fixed percentage

Money left in a bank at compound interest grows by the same percentage each year. So the yearly amounts form a GP.

Arjun puts ₹$10000$ in a deposit paying $8\%$ a year, added once a year. Each year the amount is multiplied by $1.08$:

After $n$ years he has ₹$10000 \times 1.08^n$. After $5$ years that is about ₹$14693.28$.

Decay works the same way with a ratio below $1$. A machine worth ₹$50000$ loses $20\%$ of its value every year. Its values are $50000, 40000, 32000, \ldots$, with $r = 0.8$. After $3$ years it is worth $50000 \times 0.8^3 =$ ₹$25600$.

A town of $40000$ people growing by $5\%$ a year has $40000 \times 1.05^2 = 44100$ people after $2$ years.

Practise with new numbers

Try it in Python

The program keeps the terms as exact fractions with fractions.Fraction, checks that every ratio is the same, and finds the 6th term by the formula.

Common mistakes

• Using $a r^n$ for the nth term. The first term is $a r^0$, so term $n$ is $a r^{n - 1}$.
• Finding r by subtracting. In a GP you divide one term by the one before.
• Forgetting a negative ratio. $r^3 = -8$ gives $r = -2$, and $r^2 = 9$ gives $r = 3$ or $r = -3$.
• Adding the percentage each year. Compound growth multiplies by $1 + \tfrac{p}{100}$ every year.

Remember these

• A GP has a fixed ratio $r = \dfrac{a_{k + 1}}{a_k}$, and no term is $0$.
• The nth term is $a_n = a r^{n - 1}$.
• Divide two given terms to cancel $a$ and find $r$.
• Three numbers in GP with a known product: write $\dfrac{a}{r}$, $a$, $ar$.
• $a$, $b$, $c$ are in GP exactly when $b^2 = ac$.
• Growth or decay by $p\%$ a year multiplies by $1 \pm \tfrac{p}{100}$ each year.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the common ratio of $2, -6, 18, -54, \ldots$

2. Find the 5th term of $3, 6, 12, \ldots$

3. Is $3, 6, 9$ a geometric progression? Give a reason.

4. For which $x$ are $4$, $x$ and $9$ in GP?

5. A car worth ₹$80000$ loses $10\%$ of its value each year. What is its value after $2$ years?