IB Analysis and Approaches HL · Arithmetic and geometric sequences · Section 2 of 3

Geometric sequences and infinite sums

3 worked examples · By Álvaro López

Key idea

A geometric sequence is a list of numbers in which each one is obtained by multiplying the previous one by a fixed amount, the common ratio. With two formulas, $u_{n} = u_{1} r^{n - 1}$ and $S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1}$ , you can solve almost every question; and if $∣ r ∣ < 1$ , the sum of infinitely many terms has a finite value, $S_{\infty} = \frac{u _{1}}{1 - r}$ . It is a common topic in Paper 1 and Paper 2, often combined with an arithmetic sequence.

Common ratio and general term

In an arithmetic sequence you add the same amount to get from one term to the next. In a geometric sequence you multiply. For example, $5, 10, 20, 40, \dots$ is obtained by multiplying by $2$ , and $81, 27, 9, 3, \dots$ by $\frac{1}{3}$ .

If $r > 1$ and $u_{1} > 0$ , the terms grow; if $0 < r < 1$ , they approach $0$ ; if $r < 0$ , the signs alternate. So, to find $r$ , simply divide any two consecutive terms and then check that the quotient repeats.

To reach a distant term you do not need to calculate all the previous ones. Starting from $u_{1}$ , multiply by $r$ once to reach $u_{2}$ , twice to reach $u_{3}$ , and so on: to reach $u_{n}$ you multiply $n - 1$ times.

The second form is the most convenient when you are given two terms that are not the first: from $u_{5}$ to $u_{8}$ you must multiply by $r$ three times, that is, $u_{8} = u_{5} r^{3}$ . If you are given two terms, divide them and you will obtain a power of $r$ .

Sum of the first $n$ terms

To add $S_{n} = u_{1} + u_{2} + \dots + u_{n}$ there is a trick worth understanding, because it explains the formula. Write the sum and, underneath, the same sum multiplied by $r$ :

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

When you subtract, all the middle terms cancel and only the first term above and the last term below remain:

$r S_{n} - S_{n} = u_{1} r^{n} - u_{1} ⟹ S_{n} (r - 1) = u_{1} (r^{n} - 1)$

If $r \neq = 1$ you can divide by $r - 1$ .

The two forms are identical. Use the first when $r > 1$ and the second when $r < 1$ , so that you work with positive numbers. The formula needs $u_{1}$ , $r$ and $n$ : the number of terms you add. If you start adding at a term that is not the first, first work out its value and use it as $u_{1}$ (or subtract two sums: $u_{k + 1} + \dots + u_{n} = S_{n} - S_{k}$ ).

Finding $n$ with logarithms

When the unknown is the number of terms, it appears in an exponent: for example, $u_{n} = u_{1} r^{n - 1} = B$ or $S_{n} = B$ . Isolate the power and take logarithms, as in logarithms and their properties:

$u_{1} r^{n - 1} = B ⟺ r^{n - 1} = \frac{B}{u _{1}} ⟺ (n - 1) ln r = ln \frac{B}{u _{1}} ⟺ n = 1 + \frac{ln ( B / u _{1} )}{ln r}$

As $n$ must be an integer, pay attention to what the question asks.

If it asks when a value is reached exactly and $n$ is not an integer, that value is not a term of the sequence.
If it asks when a value is exceeded, round up to the next integer when the terms are increasing.
When you divide both sides of an inequality by $ln r$ or by $ln \frac{1}{r}$ , check the sign: if $0 < r < 1$ , then $ln r < 0$ and the inequality sign reverses.

Infinite sum

What happens if you keep on adding forever? Look at $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ . Imagine a segment of length $1$ : take its half, then half of what remains, then half of what remains again. After each step, what is missing to reach $1$ is exactly the last piece you added, and that piece gets smaller and smaller. The sum approaches $1$ without ever going past it.

The same can be seen in the formula. With $r = \frac{1}{2}$ and $u_{1} = \frac{1}{2}$ : $S_{n} = \frac{1}{2} \cdot \frac{1 - ( 1/2 ) ^{n}}{1/2} = 1 - (\frac{1}{2})^{n}$ , and the power $(\frac{1}{2})^{n}$ approaches $0$ as $n$ increases.

In general, in $S_{n} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ everything depends on what $r^{n}$ does as $n$ grows:

If $∣ r ∣ < 1$ , $r^{n}$ becomes as small as you like, approaching $0$ .
If $∣ r ∣ \geq 1$ , $r^{n}$ does not approach $0$ (it grows very large, is always $1$ , or alternates between $1$ and $- 1$ ), and the sum does not settle on any number.

The series is said to converge to $\frac{u _{1}}{1 - r}$ when $∣ r ∣ < 1$ . This is the same as $- 1 < r < 1$ ; if the question says that “the infinite sum exists”, that already gives you the condition $- 1 < r < 1$ , which can help you rule out solutions.

Recurring decimals

A recurring decimal is the infinite sum of a geometric sequence with common ratio $\frac{1}{1 0 ^{k}}$ , where $k$ is the length of the repeating block. Write each piece of the repeating block as a term and apply the formula.

For example, $0. 6 = 0.6 + 0.06 + 0.006 + \dots$ has $u_{1} = 0.6$ and $r = 0.1$ , so

$0. 6 = \frac{0.6}{1 - 0.1} = \frac{0.6}{0.9} = \frac{2}{3}$

If the number has a non-recurring part, separate it: $0.1 6 = 0.1 + 0.0 6 = \frac{1}{10} + \frac{1}{10} \cdot \frac{2}{3} = \frac{1}{10} + \frac{1}{15} = \frac{1}{6}$ .

An arithmetic and a geometric sequence

A common type of problem gives an arithmetic and a geometric sequence that are related (for example, three terms of an arithmetic sequence that are consecutive terms of a geometric sequence, or two sequences with the same first term). The key is to write each term using the formula of its sequence, $u_{n} = u_{1} + (n - 1) d$ for the arithmetic one and $v_{n} = v_{1} r^{n - 1}$ for the geometric one, and to equate what the question says coincides. If the three terms are consecutive in the geometric sequence, use $b^{2} = a c$ . Solve the resulting system and, before going on, check the solutions against the data (for example, $d \neq = 0$ if the question requires it).

Worked examples

Example 1 Basic

Common ratio, general term and sum

In a geometric sequence, $u_{2} = 12$ and $u_{5} = 96$ .

a) Find the common ratio $r$ and the first term $u_{1}$ .

b) Calculate $u_{10}$ .

c) Calculate the sum of the first eight terms.

d) Find the smallest value of $n$ for which $u_{n} > 10000$ .

Solution

a) From $u_{2}$ to $u_{5}$ you multiply by $r$ three times:

$u_{5} = u_{2} r^{3} ⟹ r^{3} = \frac{96}{12} = 8 ⟹ r = 2$

Since $u_{2} = u_{1} r$ , it follows that $u_{1} = \frac{12}{2} = 6$ .

b) $u_{10} = 6 \cdot 2^{9} = 6 \cdot 512 = 3072$ .

c) Since $r > 1$ , we use $S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1}$ :

$S_{8} = \frac{6 ( 2 ^{8} - 1 )}{2 - 1} = 6 \cdot 255 = 1530$

d) We set up $6 \cdot 2^{n - 1} > 10000$ , that is, $2^{n - 1} > \frac{5000}{3} \approx 1666.7$ . Taking logarithms (with $ln 2 > 0$ ):

$n - 1 > \frac{ln ( 5000/3 )}{ln 2} \approx 10.70 ⟹ n > 11.70$

The smallest integer is $n = 12$ . Check: $u_{11} = 6 \cdot 2^{10} = 6144 < 10000$ and $u_{12} = 6 \cdot 2^{11} = 12288 > 10000$ .

Example 2 Medium

Infinite sum and recurring decimals

a) Write $0. 27 = 0.272727 \dots$ as the sum of an infinite geometric sequence and express it as a fraction in its simplest form.

b) Calculate the infinite sum $12 - 4 + \frac{4}{3} - \frac{4}{9} + \dots$

c) A geometric sequence has an infinite sum equal to $18$ and its second term is $4$ . Find the possible values of the common ratio $r$ .

Solution

a) The repeating block has two digits: $0. 27 = 0.27 + 0.0027 + 0.000027 + \dots$ , with $u_{1} = 0.27$ and $r = 0.01$ . Since $∣ r ∣ < 1$ :

$S_{\infty} = \frac{0.27}{1 - 0.01} = \frac{0.27}{0.99} = \frac{27}{99} = \frac{3}{11}$

b) The first term is $u_{1} = 12$ and, dividing, $r = \frac{- 4}{12} = - \frac{1}{3}$ (check: $- 4 \cdot (- \frac{1}{3}) = \frac{4}{3}$ ). Since $∣ r ∣ = \frac{1}{3} < 1$ , the sum exists:

$S_{\infty} = \frac{12}{1 + \frac{1}{3}} = \frac{12}{4/3} = 9$

c) The second term is $u_{2} = u_{1} r = 4$ , so $u_{1} = \frac{4}{r}$ . With $S_{\infty} = \frac{u _{1}}{1 - r} = 18$ :

$\frac{4}{r ( 1 - r )} = 18 ⟹ 9 r (1 - r) = 2 ⟹ 9 r^{2} - 9 r + 2 = 0$

$r = \frac{9 \pm 81 - 72}{18} = \frac{9 \pm 3}{18} ⟹ r = \frac{2}{3} or r = \frac{1}{3}$

Both values satisfy $∣ r ∣ < 1$ , so both are valid. Check: with $r = \frac{1}{3}$ , $u_{1} = 12$ and $S_{\infty} = \frac{12}{2/3} = 18$ ; with $r = \frac{2}{3}$ , $u_{1} = 6$ and $S_{\infty} = \frac{6}{1/3} = 18$ .

Example 3 Hard

Three terms of an arithmetic sequence that form a geometric sequence

An arithmetic sequence has first term $u_{1} = 3$ and common difference $d \neq = 0$ . Its second, fifth and fourteenth terms are, in this order, three consecutive terms of a geometric sequence.

a) Find the common difference $d$ of the arithmetic sequence and the common ratio $r$ of the geometric sequence.

b) The geometric sequence has as its first term the second term of the arithmetic sequence, $v_{1} = u_{2}$ , and the common ratio you have found. Calculate the smallest number of terms $n$ for which its sum exceeds $1 0^{6}$ .

c) Calculate the sum of the first twenty terms of the arithmetic sequence.

Solution

a) 1. Write the three terms using $u_{n} = 3 + (n - 1) d$ :

$u_{2} = 3 + d u_{5} = 3 + 4 d u_{14} = 3 + 13 d$

Impose the geometric condition $b^{2} = a c$ for three consecutive terms:

$(3 + 4 d)^{2} = (3 + d) (3 + 13 d)$

Expand and simplify:

$9 + 24 d + 16 d^{2} = 9 + 42 d + 13 d^{2} ⟹ 3 d^{2} - 18 d = 0 ⟹ 3 d (d - 6) = 0$

Reject $d = 0$ because the question requires $d \neq = 0$ . This leaves $d = 6$ .
The three terms are $u_{2} = 9$ , $u_{5} = 27$ , $u_{14} = 81$ , and the common ratio is $r = \frac{27}{9} = 3$ (also $\frac{81}{27} = 3$ , so it is consistent).

b) The geometric sequence is $v_{n} = 9 \cdot 3^{n - 1}$ and its sum is

$S_{n} = \frac{9 ( 3 ^{n} - 1 )}{3 - 1} = \frac{9 ( 3 ^{n} - 1 )}{2}$

The question asks for $\frac{9 ( 3 ^{n} - 1 )}{2} > 1 0^{6}$ , that is, $3^{n} > \frac{2 \cdot 1 0 ^{6}}{9} + 1 \approx 222 223.2$ . Using logarithms ( $ln 3 > 0$ ):

$n > \frac{ln ( 222 223.2 )}{ln 3} \approx 11.21$

The smallest integer is $n = 12$ . Check: $S_{11} = 797157 < 1 0^{6}$ and $S_{12} = 2391480 > 1 0^{6}$ .

c) With $u_{1} = 3$ and $d = 6$ :

$S_{20} = \frac{20}{2} (2 \cdot 3 + 19 \cdot 6) = 10 \cdot 120 = 1200$

In the exam: you earn marks for setting up $b^{2} = a c$ with the terms correctly written and for rejecting $d = 0$ , justifying it from the question. In b) it is not enough to give $11.21$ : you must conclude with $n = 12$ and, if you have time, check with $S_{11}$ and $S_{12}$ .

Common mistakes

Summary

Geometric sequence: $u_{n + 1} = r u_{n}$ , with common ratio $r = \frac{u _{n + 1}}{u _{n}}$ . General term: $u_{n} = u_{1} r^{n - 1}$ (also $u_{n} = u_{k} r^{n - k}$ ).
Three consecutive terms $a, b, c$ satisfy $b^{2} = a c$ .
Sum of $n$ terms, with $r \neq = 1$ : $S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ ; if $r = 1$ , $S_{n} = n u_{1}$ . The formula is obtained by subtracting $S_{n}$ and $r S_{n}$ .
To find $n$ in $u_{n} = B$ or $S_{n} = B$ : isolate the power $r^{\dots}$ , take logarithms and bring down the exponent. Round according to what the question asks and watch the sign of $ln r$ in inequalities.
If $∣ r ∣ < 1$ : $S_{\infty} = \frac{u _{1}}{1 - r}$ . If $∣ r ∣ \geq 1$ , the infinite sum does not exist.
A recurring decimal is an infinite sum with $r = 1 0^{- k}$ ( $k$ digits in the repeating block): $0. 27 = \frac{0.27}{0.99} = \frac{3}{11}$ .
In mixed problems, write each term with its formula (arithmetic or geometric), equate what the question says coincides and check the solutions.

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Geometric sequences and infinite sums

Common ratio and general term

Sum of the first n terms

Finding n with logarithms

Infinite sum

Recurring decimals

An arithmetic and a geometric sequence

Worked examples

Common mistakes

Summary

Want to work on it together?

Sum of the first $n$ terms

Finding $n$ with logarithms