AI HL · 12.1
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Geometric Series
12.1.5
Choose arithmetic or geometric, then change the first term and d or r. Before you tick the box, predict what the sums Sₙ will look like.
| n | uₙ | Sₙ |
|---|
Move a slider to change the sequence.
A geometric sequence has first term 6 and common ratio 2. Find the sum of the first 10 terms.
Have a go first. Then open the working one line at a time.
\(S_{10} = \frac{6(2^{10} - 1)}{2 - 1} = 6138\)
\(u_1 = 6\), \(r = 2\), \(n = 10\).
A geometric sequence has first term 81 and common ratio \(-\frac{1}{3}\). Find the sum of the first 6 terms, to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(S_6 = \frac{81\left(1 - \left(-\frac{1}{3}\right)^6\right)}{1 - \left(-\frac{1}{3}\right)} = 60.66\ldots\)
The ratio goes in brackets both times.
\(60.7\)
The terms are 81, −27, 9, −3, 1, \(-\frac{1}{3}\).
A geometric sequence has first term 200 and common ratio 0.5. Find the sum of the first 8 terms, to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(S_8 = \frac{200(1 - 0.5^8)}{1 - 0.5} = 398.4\ldots\)
\(r < 1\), so use the \(1 - r\) form.
\(398\)
Each term is half the last, so the sum closes in on 400.
How many terms of \(3 + 6 + 12 + \ldots\) are needed for the sum to first exceed 5000?
Have a go first. Then open the working one line at a time.
\(S_n = \frac{3(2^n - 1)}{2 - 1} = 3(2^n - 1)\)
\(u_1 = 3\), \(r = 2\).
\(S_{10} = 3069\), \(S_{11} = 6141\)
From a table on the calculator. 11 terms.
A runner covers 2 km in the first week of training and 20% more each week than the week before. Find the total distance run in the first 10 weeks.
Have a go first. Then open the working one line at a time.
\(2 + 2(1.2) + 2(1.2)^2 + \ldots\)
20% more each week multiplies by 1.2. The total is a sum, so it is a series.
\(S_{10} = \frac{2(1.2^{10} - 1)}{1.2 - 1} = 51.91\ldots\)
\(u_1 = 2\), \(r = 1.2\), \(n = 10\).
About 51.9 km
Ten weeks of running.
A new café serves 800 customers in its first month. The number of customers grows by 3% each month. Find the total number of customers in the café's second year, to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(u_1 = 800\), \(r = 1.03\)
3% more each month multiplies by 1.03. A total is a sum, so it is a series.
\(S_{24} = \frac{800(1.03^{24} - 1)}{1.03 - 1} = 27\,541.2\ldots\)
The first 24 months.
\(S_{12} = \frac{800(1.03^{12} - 1)}{1.03 - 1} = 11\,353.6\ldots\)
The first 12 months.
\(S_{24} - S_{12} = 16\,187.6\ldots\)
Months 13 to 24. About 16 200 customers. Keep the full values on the calculator and round only at the end.
A drone is tested along a straight track. Its moves, in metres, form a geometric sequence: 8, −6, 4.5, … A negative move is a move backwards. Find the total distance the drone travels in its first 6 moves, to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(r = \frac{-6}{8} = -0.75\)
The minus sign only shows the direction of each move.
8, 6, 4.5, …
Distance adds the sizes, whichever way the drone moves. They form a geometric sequence with \(u_1 = 8\) and \(r = 0.75\).
\(S_{6} = \frac{8(1 - 0.75^{6})}{1 - 0.75} = 26.30\ldots\)
About 26.3 m. With \(r = -0.75\) the sum is 3.76, which is where the drone ends up, not how far it has travelled.
A ball is dropped from a height of 4 m. After each bounce it rises to 75% of the height it fell from. Find the total distance the ball has travelled when it hits the ground for the 5th time.
Have a go first. Then open the working one line at a time.
4 m
The ball falls 4 m to the ground the 1st time.
\(3,\ 2.25,\ 1.6875,\ 1.265625\)
Four rebounds happen before the 5th hit. Each is \(r = 0.75\) of the one before: a geometric sequence.
\(2 \times \dfrac{3(1 - 0.75^4)}{1 - 0.75} = 16.4062\)
Each rebound height counts twice.
\(4 + 16.4062 = 20.4062\ldots\)
About 20.4 m.
A video has 50 views on day 1. The number of new views doubles each day. On which day does the total number of views first pass 100 000?
Have a go first. Then open the working one line at a time.
\(50 + 100 + 200 + \ldots\)
\(u_1 = 50\) and \(r = 2\). The total is a sum.
\(S_n = \dfrac{50(2^n - 1)}{2 - 1} = 50(2^n - 1)\)
Put this in a table, or solve \(S_n = 100\,000\) on the calculator.
\(S_{10} = 51 150\), \(S_{11} = 102 350\)
The total first passes 100 000 on day 11.