Adding the Terms of a Geometric Sequence
Look at the sequence 5, 15, 45, 135, 405, 1215. Each term is 3 times the term before it, so it is a geometric sequence with common ratio 3. If we join its terms with plus signs, we get the series 5 + 15 + 45 + 135 + 405 + 1215.
We can add these six terms by hand. 5 + 15 = 20. Then 20 + 45 = 65, 65 + 135 = 200, 200 + 405 = 605 and 605 + 1215 = 1820. So the sum of this series is 1820.
Six terms are easy to add. Now imagine the same series with twenty terms. The twentieth term is \( 5 \times 3^{19} \), which is more than five billion. Adding twenty numbers like that by hand is slow, and one slip spoils the answer. So we want a formula that gives the sum straight from the first term, the common ratio and the number of terms.
Geometric series: the expression we get by adding the terms of a geometric sequence (GP). For example, 5 + 15 + 45 + 135 is a geometric series. The value we get by adding its terms is called the sum of the geometric series.
Sum of n terms: the sum of the first n terms of a geometric series is written \( s_n \). So \( s_1 \) is the first term alone, \( s_2 \) is the first two terms added, and \( s_6 \) is the first six terms added.
Throughout this topic we use these letters:
- \( a \) = the first term
- \( r \) = the common ratio, found by dividing any term by the term just before it
- \( n \) = the number of terms
- \( t_n \) or \( l \) = the last term, where \( l = ar^{n-1} \)
- \( s_n \) = the sum of the first n terms
Deriving the Sum Formula
The idea is simple. We write the series out, multiply the whole series by r, and subtract one from the other. Almost every term cancels, and what is left is short enough to solve for \( s_n \).
Step 1: Writing the sum of n terms using a and r. We call this equation (i):
\[ s_n = a + ar + ar^2 + \cdots + ar^{n-1} \]
Step 2: Multiplying every term of (i) by r. Each power of r goes up by one. We call this equation (ii):
\[ rs_n = ar + ar^2 + \cdots + ar^{n-1} + ar^n \]
Now write (ii) under (i), moved one place to the right. Every term of (ii) except the last one sits exactly under an equal term of (i).
Step 3: Subtracting (ii) from (i). The terms \( ar, ar^2, \ldots, ar^{n-1} \) appear in both, so they cancel. Only the first term of (i) and the last term of (ii) remain:
\[ s_n - rs_n = a - ar^n \]
Step 4: Taking out the common factor \( s_n \) on the left and \( a \) on the right:
\[ s_n(1 - r) = a(1 - r^n) \]
Step 5: Dividing both sides by \( (1 - r) \), which is allowed because r is not 1:
\[ s_n = \frac{a(1 - r^n)}{1 - r} \]
Step 6: Multiplying the numerator and the denominator by \( -1 \). This changes nothing in the value but turns both brackets round:
\[ s_n = \frac{a(r^n - 1)}{r - 1} \]
Choosing Between the Two Forms
We now have two forms of the same formula. The first is used when r < 1, and the second when r > 1. The reason is only convenience: each choice keeps the numerator and the denominator positive, so fewer minus signs get in the way.
| Form | Used when | Example of r |
| \( s_n = \frac{a(1 - r^n)}{1 - r} \) | r < 1 | \( \frac{1}{3} \), \( \frac{1}{2} \) |
| \( s_n = \frac{a(r^n - 1)}{r - 1} \) | r > 1 | 2, 3, 4 |
The two forms are the same expression written two ways. For any value of r except 1, both give exactly the same answer. If you use the "wrong" one, you will still be right, only with more negative signs to handle.
Neither form works when r = 1, because the denominator becomes 0. In that case every term equals a, as in 4 + 4 + 4 + 4 + 4, so simply \( s_n = na \).
Let us test the formula on the series we added by hand. For 5 + 15 + 45 + 135 + 405 + 1215 we have a = 5, r = 3 and n = 6. Since r > 1, we use the second form:
\[ s_6 = \frac{5(3^6 - 1)}{3 - 1} = \frac{5 \times 728}{2} \]
\[ s_6 = 1820 \]
This is exactly the sum we found by adding, so the formula agrees with the long way.
The Sum in Terms of the Last Term
Sometimes a question gives the last term l but not the number of terms n. We can rewrite the formula so that it uses l instead of n, using the fact that \( l = ar^{n-1} \).
Step 1: Starting from the form for r > 1 and opening the bracket in the numerator:
\[ s_n = \frac{ar^n - a}{r - 1} \]
Step 2: Splitting \( ar^n \) as \( ar^{n-1} \times r \), since the powers add up to n:
\[ s_n = \frac{ar^{n-1} \times r - a}{r - 1} \]
Step 3: Replacing \( ar^{n-1} \) by the last term l:
\[ s_n = \frac{lr - a}{r - 1} \]
The same steps, started from the form for r < 1, give the matching formula for a ratio less than 1:
\[ s_n = \frac{a - lr}{1 - r} \]
The last-term forms do not contain n at all. When a question tells you the first term, the last term and the ratio, you can find the sum without first working out how many terms there are.
Choosing Unknown Terms Wisely
Many questions give the sum or the product of a few numbers in a GP and ask for the numbers. The work becomes much easier if we name the unknown terms symmetrically, with the middle of the group as the centre. Then the powers of r cancel when the terms are multiplied.
| Number of terms | Take them as | Product |
| Three | \( \frac{a}{r}, a, ar \) | \( a^3 \) |
| Four | \( \frac{a}{r^3}, \frac{a}{r}, ar, ar^3 \) | \( a^4 \) |
| Five | \( \frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2 \) | \( a^5 \) |
With three or five terms, there is a middle term a and the ratio between neighbours is r. With four terms there is no single middle term, so the two middle terms are \( \frac{a}{r} \) and \( ar \). The ratio between neighbours is then \( r^2 \), because \( ar \div \frac{a}{r} = r^2 \). In every case, a product condition gives a straight away.
Worked Example 1: Sum of a Fixed Number of Terms
Find the sum of the series 36 + 12 + 4 + ... up to 6 terms.
Step 1: Reading off the values. The first term is a = 36 and the common ratio is \( r = \frac{12}{36} = \frac{1}{3} \), so the series is geometric, with n = 6. Since r < 1, we use:
\[ s_n = \frac{a(1 - r^n)}{1 - r} \]
Step 2: Substituting a = 36, \( r = \frac{1}{3} \) and n = 6:
\[ s_6 = \frac{36\left\{1 - \left(\frac{1}{3}\right)^6\right\}}{1 - \frac{1}{3}} \]
Step 3: Working out the power and the denominator. Since \( \left(\frac{1}{3}\right)^6 = \frac{1}{729} \), the bracket is \( \frac{728}{729} \) and the denominator is \( \frac{2}{3} \):
\[ s_6 = \frac{36 \times \frac{728}{729}}{\frac{2}{3}} \]
Step 4: Dividing by \( \frac{2}{3} \), which is the same as multiplying by \( \frac{3}{2} \):
\[ s_6 = \frac{36 \times 728 \times 3}{729 \times 2} \]
Step 5: Cancelling. Since \( \frac{36 \times 3}{2} = 54 \) and \( \frac{54}{729} = \frac{2}{27} \), the sum becomes:
\[ s_6 = \frac{2 \times 728}{27} = \frac{1456}{27} = 53\frac{25}{27} \]
Check: the six terms are 36, 12, 4, \( \frac{4}{3} \), \( \frac{4}{9} \) and \( \frac{4}{27} \). The first three add to 52 and the last three add to \( \frac{52}{27} = 1\frac{25}{27} \), giving \( 53\frac{25}{27} \) again.
Worked Example 2: Finding the Common Ratio
The first and last terms of a GP are 7 and 448, and the sum of the series is 889. Find the common ratio.
Step 1: Choosing the formula. We know a = 7, l = 448 and \( s_n = 889 \), but not n. So we use the last-term form:
\[ s_n = \frac{lr - a}{r - 1} \]
Step 2: Substituting the known values:
\[ 889 = \frac{448r - 7}{r - 1} \]
Step 3: Multiplying both sides by \( (r - 1) \) to clear the fraction:
\[ 889r - 889 = 448r - 7 \]
Step 4: Collecting the r terms on the left and the numbers on the right:
\[ 889r - 448r = 889 - 7 \]
\[ 441r = 882 \]
Step 5: Dividing both sides by 441:
\[ r = \frac{882}{441} = 2 \]
So the common ratio is 2. Check: the series is 7 + 14 + 28 + 56 + 112 + 224 + 448, and these seven terms add up to 889.
Worked Example 3: Using Two Sums
The sum of the first two terms of a GP is 6 and the sum of the first four terms is 30. Find the sum of the first 10 terms.
Step 1: Writing \( s_2 = 6 \) with the formula. We call this equation (i):
\[ 6 = \frac{a(r^2 - 1)}{r - 1} \]
Step 2: Writing \( s_4 = 30 \) and factorising \( r^4 - 1 = (r^2)^2 - 1^2 \) as a difference of two squares. We call this equation (ii):
\[ 30 = \frac{a(r^2 + 1)(r^2 - 1)}{r - 1} \]
Step 3: Dividing (ii) by (i). The whole of \( \frac{a(r^2 - 1)}{r - 1} \) appears in both, so it cancels:
\[ \frac{30}{6} = r^2 + 1 \]
Step 4: Solving for r. Since 5 = \( r^2 + 1 \):
\[ r^2 = 4 \]
\[ r = \pm 2 \]
Both values are possible, so we work each case separately.
Step 5: Case I, putting r = 2 into equation (i):
\[ 6 = \frac{a(2^2 - 1)}{2 - 1} = 3a \]
So a = 2. Using a = 2, r = 2 and n = 10:
\[ s_{10} = \frac{2(2^{10} - 1)}{2 - 1} \]
\[ s_{10} = 2 \times 1023 = 2046 \]
Step 6: Case II, putting r = \( -2 \) into equation (i):
\[ 6 = \frac{a\{(-2)^2 - 1\}}{-2 - 1} = \frac{3a}{-3} \]
So \( -a = 6 \) and a = \( -6 \). Using a = \( -6 \), r = \( -2 \) and n = 10:
\[ s_{10} = \frac{-6\{(-2)^{10} - 1\}}{-2 - 1} \]
\[ s_{10} = \frac{-6 \times 1023}{-3} = 2046 \]
Both cases give the same answer, so the sum of the first 10 terms is 2046.
Worked Example 4: Three Numbers with a Given Sum and Product
The sum of three numbers in geometric progression is 52 and their product is 1728. Find the numbers.
Step 1: Naming the numbers symmetrically, as \( \frac{a}{r} \), a and ar. Using the product condition first, because r cancels there:
\[ \frac{a}{r} \times a \times ar = 1728 \]
\[ a^3 = 1728 \]
Step 2: Taking the cube root of both sides:
\[ a = \sqrt[3]{1728} = 12 \]
Step 3: Putting a = 12 into the sum condition:
\[ \frac{12}{r} + 12 + 12r = 52 \]
Step 4: Multiplying every term by r to remove the fraction:
\[ 12 + 12r + 12r^2 = 52r \]
Step 5: Dividing by 4 and bringing every term to one side:
\[ 3r^2 - 10r + 3 = 0 \]
Step 6: Factorising by splitting the middle term, since \( -10r = -9r - r \):
\[ 3r^2 - 9r - r + 3 = 0 \]
\[ 3r(r - 3) - 1(r - 3) = 0 \]
\[ (r - 3)(3r - 1) = 0 \]
So either r = 3 or \( r = \frac{1}{3} \).
Step 7: Writing the numbers for each value of r. When r = 3, the numbers are \( \frac{12}{3} = 4 \), 12 and \( 12 \times 3 = 36 \). When \( r = \frac{1}{3} \), they are 36, 12 and 4.
So the numbers are 4, 12, 36 or 36, 12, 4. Check: 4 + 12 + 36 = 52 and \( 4 \times 12 \times 36 = 1728 \).
Worked Example 5: Finding the Number of Terms
How many terms of the series 3 + 6 + 12 + ... must be taken so that the sum is 381?
Step 1: Here a = 3, r = 2 and \( s_n = 381 \), with n unknown. Since r > 1, we substitute into the second form:
\[ 381 = \frac{3(2^n - 1)}{2 - 1} \]
Step 2: The denominator is 1, so dividing both sides by 3:
\[ 2^n - 1 = 127 \]
Step 3: Adding 1 to both sides and writing 128 as a power of 2:
\[ 2^n = 128 = 2^7 \]
Step 4: The bases are equal, so the powers are equal:
\[ n = 7 \]
So 7 terms are needed. Check: 3 + 6 + 12 + 24 + 48 + 96 + 192 = 381.
Worked Example 6: Sum When the Last Term Is Given
Find the sum of 1 + 4 + 16 + ... + 4096.
Step 1: Here a = 1, r = 4 and the last term is l = 4096. Since the last term is given and r > 1, we use:
\[ s_n = \frac{lr - a}{r - 1} \]
Step 2: Substituting l = 4096, r = 4 and a = 1:
\[ s_n = \frac{4096 \times 4 - 1}{4 - 1} \]
Step 3: Multiplying and then dividing:
\[ s_n = \frac{16383}{3} = 5461 \]
We never needed to know that there are 7 terms. That is exactly why the last-term form is useful.
All the Sum Formulas Together
| What you know | Formula for the sum |
| a, r, n with r < 1 | \( \frac{a(1 - r^n)}{1 - r} \) |
| a, r, n with r > 1 | \( \frac{a(r^n - 1)}{r - 1} \) |
| a, r, l with r > 1 | \( \frac{lr - a}{r - 1} \) |
| a, r, l with r < 1 | \( \frac{a - lr}{1 - r} \) |
| r = 1 | \( na \) |
Every sum question uses one of these. Read the question, list what is given and what is asked, and then pick the row that uses exactly the quantities you have. When the question hides the unknown inside the formula, as in finding r, a or n, the same formula becomes an equation to solve.
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