A-level Mathematics

4.6 Geometric sequences and series, and the sum to infinity — Learn

Prerequisites: the sequences in Section 4.3; sigma notation in Section 4.4; the laws of indices from Section 2.1, including negative powers.

By the end of this section you can: find the nnth term of a geometric sequence; find the sum of a finite geometric series; find the common ratio from two terms; decide when a geometric series converges; and find its sum to infinity.

1. The common ratio

A geometric sequence is made by multiplying by the same amount at every step. That fixed multiplier is the common ratio, written rr. With first term aa,

un=arn−1.u_n = a r^{n-1}. 3,  6,  12,  24,  …3, \; 6, \; 12, \; 24, \; \dots

Here a=3a = 3 and r=2r = 2, because each term is twice the one before. A common ratio between 00 and 11 makes the sequence fall; a negative ratio makes the terms alternate in sign.

2. The sum of a finite geometric series

The sum of the first nn terms is

Sn=a(1−rn)1−r,r≠1.S_n = \frac{a(1 - r^n)}{1 - r}, \qquad r \ne 1.

Example 1. For 3,6,12,…3, 6, 12, \dots, work out the 10th term and the sum of the first 10 terms.

The 10th term is

u10=3(2)10−1=3(512)=1536.u_{10} = 3(2)^{10-1} = 3(512) = 1536.

The sum uses a=3a = 3, r=2r = 2 and n=10n = 10:

S10=3(1−210)1−2=3(1−1024)−1=3(1023)=3069.S_{10} = \frac{3(1 - 2^{10})}{1 - 2} = \frac{3(1 - 1024)}{-1} = 3(1023) = 3069.

The sum is 30693069. Notice that the sum is about twice the last term, which is what a ratio of 22 gives.

3. Finding the common ratio

Two terms are enough to find rr, because the ratio between them is a power of rr.

Example 2. A geometric sequence has first term 88 and third term 1818. Find the two possible values of the common ratio.

The third term is ar2=18a r^2 = 18, so

8r2=18  ⟹  r2=188=94,8r^2 = 18 \implies r^2 = \frac{18}{8} = \frac{9}{4}, r=32orr=−32.r = \frac{3}{2} \quad \text{or} \quad r = -\frac{3}{2}.

Both values fit the information given. A question that says the terms are positive picks the positive value.

4. The sum to infinity

When ∣r∣<1|r| < 1, each term is smaller than the one before and the running total settles on a fixed value. That value is the sum to infinity:

S∞=a1−r,∣r∣<1.S_\infty = \frac{a}{1 - r}, \qquad |r| < 1.

When ∣r∣≥1|r| \ge 1, the running total does not settle. The terms do not shrink, or they grow, so the series diverges and has no sum to infinity. The condition ∣r∣<1|r| < 1 is part of the answer, not an extra.

Example 3. The online appeal in the chapter problem raises £600 in month 1 and 80% of the previous month after that. Work out the most the appeal can ever raise.

Here a=600a = 600 and r=0.8r = 0.8, and ∣0.8∣<1|0.8| < 1, so the series converges:

S∞=6001−0.8=6000.2=3000.S_\infty = \frac{600}{1 - 0.8} = \frac{600}{0.2} = 3000.

The appeal can never raise more than £3000, however long it runs. After a year it has already collected most of that.

Example 4. Work out the exact total raised by the appeal in the first 12 months, and compare it with the sum to infinity.

Use the finite sum with n=12n = 12:

S12=600(1−0.812)1−0.8=3000(1−0.812).S_{12} = \frac{600(1 - 0.8^{12})}{1 - 0.8} = 3000(1 - 0.8^{12}).

Now 0.812=0.0687194767360.8^{12} = 0.068719476736, so

S12=3000(0.931280523264)=2793.84 (to the nearest penny).S_{12} = 3000(0.931280523264) = 2793.84 \text{ (to the nearest penny)}.

The appeal raises £2793.84 in the first year, which is £206.16 short of the £3000 it can never quite reach.

5. A recurring decimal as a geometric series

A recurring decimal is an infinite geometric series, which is why it can be written as a fraction.

Example 5. Write 0.777…0.777\dots as a fraction.

Write the decimal as a sum:

0.777⋯=0.7+0.07+0.007+…0.777\dots = 0.7 + 0.07 + 0.007 + \dots

This is geometric with a=0.7a = 0.7 and r=0.1r = 0.1, and ∣0.1∣<1|0.1| < 1, so

0.777⋯=0.71−0.1=0.70.9=79.0.777\dots = \frac{0.7}{1 - 0.1} = \frac{0.7}{0.9} = \frac{7}{9}.

6. Try it: the geometric sum explorer

Predict first. The figure draws the running total of a geometric series as more terms are added. Before you touch the control, predict what happens to the running total when r=1.2r = 1.2, and what happens when r=0.6r = 0.6. Write both predictions down.

Test. Move the slider for rr and watch the running total. Keep an eye on the line that marks the sum to infinity. The readout names the first term, the common ratio and the limit.

Explain. For r=0.6r = 0.6, describe in two sentences why the running total creeps towards one value and never passes it. Then say what changes when rr rises above 11.

Watch out. The exponent is the commonest trap.

Using rnr^n for the nnth term. The nnth term is arn−1a r^{n-1}. For 3,6,12,…3, 6, 12, \dots the 4th term is 3(2)3=243(2)^3 = 24. Writing 3(2)4=483(2)^4 = 48 has taken one step too many, because the first term is the starting value. Re-check: put n=1n = 1. The formula must give ar0=aa r^0 = a.

Using the sum to infinity when the series diverges. The formula S∞=a1−rS_\infty = \frac{a}{1 - r} needs ∣r∣<1|r| < 1. For the quiz night in the chapter problem the income rises by £50 each month, so it is not geometric at all; for 2+4+8+…2 + 4 + 8 + \dots the ratio is 22, so ∣r∣≥1|r| \ge 1 and there is no sum to infinity. Re-check: write ∣r∣<1|r| < 1 beside every use of the formula.

Progress check

Return to the chapter problem. The online appeal has a=600a = 600 and r=0.8r = 0.8.

  1. Work out the finite sum S12S_{12} with the formula for a geometric series.
  2. Work out the sum to infinity, and say what it means for the fund.
  3. The total after 12 months is £2793.84. How much of the £3000 total is left for the months after the first year?
  4. Explain why you cannot use the same formula for the quiz night income.

You should now be able to

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