IB Applications and Interpretation SL · Arithmetic and geometric sequences · Section 3 of 5

Sum of a geometric progression

3 worked examples · By Álvaro López

Key idea

Adding the terms of a geometric progression one by one is slow when there are many of them, and a formula does it in one step. In this section you derive it, use it with a ratio other than 1 and with ratio 1, and learn how to find how many terms are needed to reach a given total, using logarithms or the graphic display calculator (GDC). It is the calculation behind increasing monthly savings, salaries with percentage rises and many Paper 2 problems.

What is being summed

Remember that in a geometric progression each term is obtained by multiplying the previous one by the ratio $r$ , so $u_{n} = u_{1} r^{n - 1}$ . The sum of the first $n$ terms is written $S_{n}$ :

$S_{n} = u_{1} + u_{2} + \dots + u_{n} = u_{1} + u_{1} r + u_{1} r^{2} + \dots + u_{1} r^{n - 1}$

A sum of this type is called a geometric series. Notice that there are $n$ summands, and the last one has exponent $n - 1$ , not $n$ .

The formula

The idea behind the proof is a trick: multiply the sum by $r$ and subtract the original. Almost everything cancels.

$S_{n} r S_{n} = u_{1} + u_{1} r + u_{1} r^{2} + \dots + u_{1} r^{n - 1} = u_{1} + u_{1} r + u_{1} r^{2} + \dots + u_{1} r^{n - 1} + u_{1} r^{n}$

Subtracting, only the first term of the top row and the last term of the bottom row survive: $r S_{n} - S_{n} = u_{1} r^{n} - u_{1}$ . Taking out a common factor, $S_{n} (r - 1) = u_{1} (r^{n} - 1)$ , and dividing by $r - 1$ (which is why $r \neq = 1$ is needed) gives the formula.

When the ratio is 1

If $r = 1$ all the terms are equal to $u_{1}$ and there is nothing to derive: adding the same number $n$ times is multiplication. In that case the formula above cannot be used, because it would have a $0$ in the denominator.

For example, the progression $4, 4, 4, \dots$ sums to $4 n$ when you take $n$ terms. In practice this case hardly appears, because a progression with ratio 1 neither grows nor decreases.

How it is applied

Finding the number of terms

Sometimes the total is given and you are asked how many terms are needed to reach or exceed it. Then $n$ is in an exponent and there are two routes.

With logarithms. Isolate the power $r^{n}$ and apply a logarithm to both sides. For $r > 1$ , from $S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1}$ we get $r^{n} = 1 + \frac{S _{n} ( r - 1 )}{u _{1}}$ , and from there $n = \frac{ln ( 1 + \frac{S _{n} ( r - 1 )}{u _{1}} )}{ln r}$ . It is a typical Paper 1 question if the numbers are simple.

Note · With the GDC

In Paper 2 you can avoid logarithms. One way is to solve the equation $S_{n} = total$ with the equation solver of the GDC, taking $n$ as the unknown. Another is to plot $y = \frac{u _{1} ( r ^{x} - 1 )}{r - 1}$ and the horizontal line $y = total$ , and read the $x$ coordinate of the intersection point. A third is to create a table of values of $S_{n}$ and see between which two values of $n$ the total is passed. The GDC can also calculate directly a sum of terms of the progression when $n$ is known, which is useful for checking the result by hand. Keep in mind that $n$ counts terms, so it must be an integer: the solution the GDC gives is often a decimal and must be rounded to the appropriate integer. If the question asks "when is the total exceeded", round up; always check by calculating $S_{n}$ with that integer and with the previous one.

Worked examples

Example 1 Basic

Summing the first terms

A geometric progression has first term $u_{1} = 5$ and ratio $r = 3$ .

a) Find the sixth term.

b) Calculate the sum of the first six terms.

c) Another progression is $7, 7, 7, \dots$ Calculate the sum of its first 20 terms.

Solution

a) $u_{6} = 5 \cdot 3^{5} = 5 \cdot 243 = 1215$ .

b) As $r > 1$ , we use the first form of the formula: $S_{6} = \frac{5 ( 3 ^{6} - 1 )}{3 - 1} = \frac{5 \cdot 728}{2} = 1820$ Check by adding: $5 + 15 + 45 + 135 + 405 + 1215 = 1820$ .

c) The ratio is $r = 1$ , so the general formula does not apply and the sum is $S_{20} = 20 \cdot 7 = 140$ .

Example 2 Medium

Downloads of an app

An app is launched and on the first day it is downloaded 200 times. Each following day it is downloaded 20% more times than the previous day.

a) State the ratio of the progression formed by the daily downloads.

b) Calculate how many downloads there are on the seventh day, rounding to the nearest unit.

c) Calculate how many downloads there are in total during the first 10 days, rounding to the nearest unit.

Solution

a) Increasing by 20% means multiplying by $1 + 0.20 = 1.2$ , so $r = \frac{6}{5} = 1.2$ and $u_{1} = 200$ .

b) $u_{7} = 200 \cdot 1. 2^{6} = 200 \cdot 2.985984 = 597.1968$ , that is, about 597 downloads.

c) We use $S_{10}$ with $r > 1$ : $S_{10} = \frac{200 ( 1. 2 ^{10} - 1 )}{1.2 - 1} = \frac{200 \cdot 5.1917364 \dots}{0.2} = 5191.7364 \dots$ In total, about 5192 downloads in the first 10 days.

Example 3 Hard

How many years to exceed the target?

A council recycles 500 tonnes of glass in the first year of a programme and, each year, recycles 10% more than the previous year.

a) Calculate the total tonnes recycled during the first 8 years, rounding to the nearest unit.

b) The target is to accumulate more than 10,000 tonnes. Using logarithms, find the minimum number of years needed.

c) Check the result of b) by calculating the total accumulated with the number of years obtained and with the previous one.

Solution

It is a geometric progression with $u_{1} = 500$ and $r = 1.1$ .

a) $S_{8} = \frac{500 ( 1. 1 ^{8} - 1 )}{0.1} = 5000 (2.14358881 - 1) = 5717.94 \dots \approx 5718 tonnes$

b) We need $S_{n} > 10000$ : $\frac{500 ( 1. 1 ^{n} - 1 )}{0.1} > 10000 \Rightarrow 5000 (1. 1^{n} - 1) > 10000 \Rightarrow 1. 1^{n} > 3$ Taking natural logarithms (since $ln 1.1 > 0$ , the direction of the inequality is preserved): $n ln 1.1 > ln 3$ , so $n > \frac{ln 3}{ln 1.1} = \frac{1.0986 \dots}{0.09531 \dots} = 11.527 \dots$ Since $n$ is an integer, the minimum is $n = 12$ years.

c) With $n = 11$ : $S_{11} = 5000 (1. 1^{11} - 1) = 5000 \cdot 1.853117 \dots = 9265.6 \dots < 10000$ . With $n = 12$ : $S_{12} = 5000 (3.138428 \dots - 1) = 10 692.1 \dots > 10000$ . Confirmed: the target is exceeded for the first time after 12 years.

In the exam: in a question like this, marks are given for the correct formula with $r = 1.1$ (not $0.1$ ), for isolating the power, for using the logarithm and, above all, for rounding $n$ up to an integer. With the GDC, write down the equation or graph you used.

Common mistakes

Summary

Geometric series: $S_{n} = u_{1} + u_{1} r + \dots + u_{1} r^{n - 1}$ , with $n$ summands.
If $r \neq = 1$ : $S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ . The formula comes from subtracting $S_{n}$ from $r S_{n}$ .
If $r = 1$ : $S_{n} = n u_{1}$ .
A percentage increase or decrease is converted into a ratio: $+ p %$ gives $r = 1 + \frac{p}{100}$ and $- p %$ gives $r = 1 - \frac{p}{100}$ .
To find $n$ : isolate $r^{n}$ and apply logarithms, or use the GDC (equation, graph or table). Round $n$ to an integer and check with $S_{n}$ .
Calculate with all the figures and round at the end, to three significant figures unless otherwise stated.

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