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Arithmetic series
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In an arithmetic sequence you add the same amount each time. Two formulae give any term and the sum of any number of terms.
The nth term
First term a, common difference d.
un = a + (n − 1)d
7, 11, 15, ...: u20 = 7 + 19 × 4 = 83
un = a + (n − 1)d
7, 11, 15, ...: u20 = 7 + 19 × 4 = 83
The sum of n terms
Sn = n2[2a + (n − 1)d]
or Sn = n2(a + l), where l is the last term.
S20 of 7, 11, 15, ... = 10(14 + 76) = 900
or Sn = n2(a + l), where l is the last term.
S20 of 7, 11, 15, ... = 10(14 + 76) = 900
Two unknowns
If u3 = 11 and u10 = 32:
a + 2d = 11 and a + 9d = 32
Subtract: 7d = 21, so d = 3 and a = 5.
a + 2d = 11 and a + 9d = 32
Subtract: 7d = 21, so d = 3 and a = 5.
How many terms of 5 + 9 + 13 + ... are needed for the sum to exceed 500?
- Sn = n2[10 + 4(n − 1)] = n(2n + 3)
- n = 15: 15 × 33 = 495, not enough
- n = 16: 16 × 35 = 560
Answer: 16 terms
un = a + (n − 1)d. Sn = n2[2a + (n − 1)d] = n2(a + l).
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