AI SL · 12.1
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Arithmetic Series
12.1.2
Using functions and tables are powerful calculator tools: put the sum in a table and read off the value of n.
Menu > Table, or a Lists & Spreadsheet page
Screenshot and video to come
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.
Find the sum of the first 30 terms of the series 5 + 9 + 13 + …
Have a go first. Then open the working one line at a time.
\(u_1 = 5\), \(d = 4\)
The terms go up by 4.
\(S_{30} = 15[2(5) + 29(4)] = 1890\)
\(S_n = \frac{n}{2}[2u_1 + (n-1)d]\) with \(n = 30\).
Find the sum of the first 20 terms of 3 + 8 + 13 + …
Have a go first. Then open the working one line at a time.
\(u_1 = 3\), \(d = 5\)
The terms go up by 5.
\(S_{20} = 10[2(3) + 19(5)] = 1010\)
\(\frac{n}{2} = 10\).
Find the sum of the first 50 terms of 100 + 93 + 86 + …
Have a go first. Then open the working one line at a time.
\(u_1 = 100\), \(d = -7\)
The terms go down by 7.
\(S_{50} = 25[2(100) + 49(-7)] = -3575\)
Later terms are negative and outweigh the early ones.
Find the sum of the first 40 positive even numbers.
Have a go first. Then open the working one line at a time.
2 + 4 + 6 + … + 80
\(u_1 = 2\), \(d = 2\), \(n = 40\).
\(S_{40} = 20(2 + 80) = 1640\)
\(S_n = \frac{n}{2}(u_1 + u_n)\) when the last term is known.
An arithmetic series has 24 terms, first term 7 and last term 122. Find its sum.
Have a go first. Then open the working one line at a time.
\(S_{24} = 12(7 + 122) = 1548\)
\(S_n = \frac{n}{2}(u_1 + u_n)\). No need to find \(d\).
How many terms of the series 3 + 7 + 11 + … are needed for the sum to first exceed 500?
Have a go first. Then open the working one line at a time.
\(S_n = \frac{n}{2}[2(3) + (n-1)(4)] = n(2n + 1)\)
\(u_1 = 3\), \(d = 4\).
\(n(2n + 1) > 500\) gives \(n = 16\)
From a table on the calculator: \(S_{15} = 465\), \(S_{16} = 528\).
How many terms of 5 + 9 + 13 + … are needed for the sum to first exceed 300?
Have a go first. Then open the working one line at a time.
\(S_n = \frac{n}{2}[2(5) + (n-1)(4)] = n(2n + 3)\)
\(u_1 = 5\), \(d = 4\).
\(S_{11} = 275\), \(S_{12} = 324\)
12 terms.
The sum of the first \(n\) terms of 2 + 5 + 8 + … is 610. Find \(n\).
Have a go first. Then open the working one line at a time.
\(S_n = \frac{n}{2}[2(2) + (n-1)(3)] = \frac{n(3n + 1)}{2}\)
\(u_1 = 2\), \(d = 3\).
\(\frac{n(3n+1)}{2} = 610\) gives \(n = 20\)
From a table, or nSolve. Check: \(\frac{20 \times 61}{2} = 610\).
For 60, 56, 52, …, how many terms are positive?
Have a go first. Then open the working one line at a time.
\(u_n = 60 + (n-1)(-4) = 64 - 4n\)
\(u_1 = 60\), \(d = -4\).
\(64 - 4n > 0\) gives \(n < 16\)
\(u_{15} = 4\) and \(u_{16} = 0\), which is not positive. 15 terms.
A worker saves $50 in week 1 and $8 more each week. In which week do total savings first pass $2000?
Have a go first. Then open the working one line at a time.
\(S_n = \frac{n}{2}[2(50) + (n-1)(8)] = n(4n + 46)\)
\(u_1 = 50\), \(d = 8\). The total is a sum.
\(S_{17} = 1938\), \(S_{18} = 2124\)
Week 18.
A cyclist rides 14 km on day 1 of a training plan and 2 km more each day than the day before. The plan lasts 25 days. Find the mean distance she rides per day.
Have a go first. Then open the working one line at a time.
\(u_{25} = 14 + (25 - 1)(2) = 62\)
24 steps of 2 from 14.
\(S_{25} = \frac{25}{2}(14 + 62) = 950\)
She rides 950 km altogether.
\(950 \div 25 = 38\)
Mean = sum ÷ number of terms. 38 km per day.
\(\frac{14 + 62}{2} = 38\)
For an arithmetic sequence the mean of the terms is also the mean of the first and last terms.
Aroha reads 12 pages of a book on day 1 and \(x\) more pages each day than the day before. She wants to read at least 1100 pages in 20 days. Find the smallest whole number value of \(x\).
Have a go first. Then open the working one line at a time.
\(S_{20} = \frac{20}{2}[2(12) + (20 - 1)x] = 240 + 190x\)
\(u_1 = 12\), \(d = x\), \(n = 20\). The total is a sum.
\(240 + 190x \ge 1100\) gives \(x \ge 4.53\ldots\)
By hand, or with nSolve on the calculator.
\(x = 5\)
\(x = 4\) gives 1000 pages, not enough. \(x = 5\) gives 1190 pages.
The 3rd term of an arithmetic series is 11 and the 8th term is 31. Find the sum of the first 20 terms.
Have a go first. Then open the working one line at a time.
\(u_1 + 2d = 11\) and \(u_1 + 7d = 31\)
Write each given term with the rule \(u_n = u_1 + (n-1)d\).
\(d = 4,\ u_1 = 3\)
Subtract: \(5d = 20\).
\(S_{20} = \frac{20}{2}\,[2(3) + 19(4)] = 820\)
Use \(S_n = \frac{n}{2}[2u_1 + (n-1)d]\).
The sum of the first \(n\) terms of a sequence is \(S_n = 2n^2 + 5n\). Find the first term and a rule for \(u_n\), and then find \(u_{12}\).
Have a go first. Then open the working one line at a time.
\(u_1 = S_1 = 2 + 5 = 7\)
The sum of one term is the term.
\(u_n = S_n - S_{n-1} = 2n^2 + 5n - [2(n-1)^2 + 5(n-1)] = 4n + 3\)
A term is the sum up to it less the sum up to the term before.
\(u_{12} = 4(12) + 3 = 51\)
Check: \(S_{12} - S_{11} = 348 - 297 = 51\).