Sums of geometric progressions

A shortcut for the sum

Find the sum of the first $8$ terms of $3, 6, 12, 24, \ldots$

Adding eight numbers works, but there is a trick that works for any GP. Write the sum, and the sum times $r$:

Subtract the first line from the second. Every term in the middle cancels:

The formula

Do the same with letters. Let $S_n = a + ar + ar^2 + \cdots + ar^{n - 1}$. Then

Subtracting, $r S_n - S_n = ar^n - a$. So, when $r \ne 1$,

Both forms are equal. The first is easier when $r > 1$, the second when $r < 1$.

When $r = 1$ every term is $a$, so $S_n = na$. For example, $7 + 7 + \cdots$ to $n$ terms is $7n$. Check $r = 1$ before you use the formula, since it would divide by $0$.

An example with r below 1

Find $1 + \tfrac{1}{2} + \tfrac{1}{4} + \cdots$ to $10$ terms.

Here $a = 1$ and $r = \tfrac{1}{2}$. So

That is just under $2$. Ten more terms would bring it closer still.

The sum to infinity

When $|r| < 1$, the power $r^n$ shrinks towards $0$ as $n$ grows. Then $S_n = \dfrac{a(1 - r^n)}{1 - r}$ gets as close as we like to

This is the sum to infinity. For example, $9 + 3 + 1 + \tfrac{1}{3} + \cdots$ has $a = 9$ and $r = \tfrac{1}{3}$, so its sum to infinity is $\dfrac{9}{1 - \tfrac{1}{3}} = \dfrac{27}{2}$.

When $|r| \ge 1$ the terms do not shrink, and there is no sum to infinity.

Set $r = -0.5$. The dots now jump above and below the red line, but they still close in on it.

Recurring decimals

A recurring decimal is a GP in disguise.

Here $a = 0.4$ and $r = 0.1$. So $0.\overline{4} = \dfrac{0.4}{0.9} = \dfrac{4}{9}$.

With two repeating digits the ratio is $\tfrac{1}{100}$. So $0.\overline{27} = \dfrac{0.27}{0.99} = \dfrac{27}{99} = \dfrac{3}{11}$.

A bouncing ball

A ball is dropped from a height of $10$ m. Each time it bounces, it rises to $\tfrac{3}{5}$ of the height it fell from. How far does it travel in all, until it comes to rest?

Step 1. The first fall. $10$ m.

Step 2. The bounces. Each bounce goes up and comes down the same height. The heights are $6, 3.6, 2.16, \ldots$, a GP with $a = 6$ and $r = \tfrac{3}{5}$.

Step 3. Add the heights. $\dfrac{6}{1 - \tfrac{3}{5}} = 15$ m. Up and down doubles this: $30$ m.

Answer: $10 + 30 = 40$ m in all.

How many terms are needed?

How many terms of $1 + 2 + 4 + \cdots$ must be added for the sum to pass $1000$?

The sum of $n$ terms is $2^n - 1$. We need $2^n - 1 > 1000$, so $2^n > 1001$. Since $2^9 = 512$ and $2^{10} = 1024$, we need $n = 10$ terms.

Savings that grow

Neha saves ₹$500$ in the first year, and $10\%$ more each year than the year before. The yearly savings form a GP with $a = 500$ and $r = 1.1$. In $4$ years she saves

that is, ₹$2320.50$.

Practise with new numbers

Try it in Python

The program adds the terms of $\tfrac{1}{2} + \tfrac{1}{4} + \cdots$ one at a time, exactly, and compares each sum with the formula and with the sum to infinity.

Common mistakes

• Using the formula when $r = 1$. It divides by $0$. The sum is $na$.
• A sum to infinity with $|r| \ge 1$. The terms do not shrink, so there is none.
• Forgetting the way back up. A bouncing ball travels each rebound height twice.
• Writing $0.\overline{27}$ as $\tfrac{27}{100}$. Two repeating digits give a ratio of $\tfrac{1}{100}$, and $\tfrac{27}{99}$.
• Using $1 + r$ for $1 - r$. The formula is $\dfrac{a}{1 - r}$, and a negative $r$ makes the bottom larger.

Remember these

• $S_n = \dfrac{a(r^n - 1)}{r - 1} = \dfrac{a(1 - r^n)}{1 - r}$ when $r \ne 1$.
• $S_n = na$ when $r = 1$.
• $S_\infty = \dfrac{a}{1 - r}$ only when $|r| < 1$.
• A recurring decimal is a GP with ratio $\tfrac{1}{10}$, $\tfrac{1}{100}$ and so on.
• To find how many terms are needed, solve the inequality for $n$.

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 sum of the first $5$ terms of $1, 4, 16, \ldots$

2. Find the sum to infinity of $8 + 4 + 2 + \cdots$

3. Write $0.\overline{8}$ as a fraction.

4. Does $1 + 3 + 9 + \cdots$ have a sum to infinity? Why?

5. Find the sum of the first $6$ terms of $5 + 5 + 5 + \cdots$