Sequences and Series
1 Arithmetic
1.1 Arithmetic sequences
An arithmetic sequence is one in which the difference between consecutive terms is constant throughout.
They are of the form:
𝑎, 𝑎 + 𝑥, 𝑎 + 2𝑥, … , 𝑎 + (𝑛 − 1)𝑥
…where 𝑎 is the value of the first term, and 𝑥 is the common difference.
Notice that the 𝑛th term is: 𝑎 + (𝑛 − 1)𝑥, as opposed to: 𝑎 + 𝑛𝑥. Consider the coefficients in front of 𝑥:
0, 1, 2, . . . , (𝑛 − 1)
There are 𝑛 coefficients, so there must be 𝑛 terms in the sequence above, hence the formula for the 𝑛th
term in a sequence is:
𝑢𝑛 = 𝑎 + (𝑛 − 1)𝑥
(Note: 0 arises from the first term – we can envisage it as: 𝑎 + 0𝑥 (= 𝑎).)
1.2 Arithmetic series
An arithmetic series is the sum of the first 𝑛 terms in an arithmetic sequence. It has the symbol: 𝑆𝑛 .
Using this definition, and (1.1), we have:
𝑆𝑛 = 𝑎 + (𝑎 + 𝑥) + (𝑎 + 2𝑥)+. . . +(𝑎 + (𝑛 − 2)𝑥) + (𝑎 + (𝑛 − 1)𝑥)
Summing this all together looks pretty tough. Instead, let’s ‘flip’ the terms around and put it side-by-side
with the original expression, in order to spot a pattern:
𝑆𝑛 = 𝑎 + (𝑎 + 𝑥) + (𝑎 + 2𝑥) + ⋯ + (𝑎 + (𝑛 − 2)𝑥) + (𝑎 + (𝑛 − 1)𝑥)
𝑆𝑛 = (𝑎 + (𝑛 − 1)𝑥) + (𝑎 + (𝑛 − 2)𝑥) + (𝑎 + (𝑛 − 2)𝑥)+. . . +(𝑎 + 𝑥) + 𝑎
‘Pair them up’ as shown above. You can see that each pair sums to: 2𝑎 + (𝑛 − 1)𝑥.
Since there are 𝑛 terms in the expression 𝑆𝑛 , then there are 𝑛 pairs in total as well.
If we sum the 2 expressions together, then, we get:
2𝑆𝑛 = 𝑛 × (2𝑎 + (𝑛 − 1)𝑥)
1
∴ 𝑆𝑛 = × 𝑛 × (2𝑎 + (𝑛 − 1)𝑥)
2
Remembering this formula would be horrendous, but fortunately we can see that the last term in the
sequence, call it 𝑙, was: 𝑙 = 𝑎 + (𝑛 − 1)𝑥.
∴ 2𝑎 + (𝑛 − 1)𝑥 = 𝑎 + (𝑎 + (𝑛 − 1)𝑥) = 𝑎 + 𝑙
…which is the first term + the last term of the series. And, substituting this into our expression for 𝑆𝑛 :
1 Arithmetic
1.1 Arithmetic sequences
An arithmetic sequence is one in which the difference between consecutive terms is constant throughout.
They are of the form:
𝑎, 𝑎 + 𝑥, 𝑎 + 2𝑥, … , 𝑎 + (𝑛 − 1)𝑥
…where 𝑎 is the value of the first term, and 𝑥 is the common difference.
Notice that the 𝑛th term is: 𝑎 + (𝑛 − 1)𝑥, as opposed to: 𝑎 + 𝑛𝑥. Consider the coefficients in front of 𝑥:
0, 1, 2, . . . , (𝑛 − 1)
There are 𝑛 coefficients, so there must be 𝑛 terms in the sequence above, hence the formula for the 𝑛th
term in a sequence is:
𝑢𝑛 = 𝑎 + (𝑛 − 1)𝑥
(Note: 0 arises from the first term – we can envisage it as: 𝑎 + 0𝑥 (= 𝑎).)
1.2 Arithmetic series
An arithmetic series is the sum of the first 𝑛 terms in an arithmetic sequence. It has the symbol: 𝑆𝑛 .
Using this definition, and (1.1), we have:
𝑆𝑛 = 𝑎 + (𝑎 + 𝑥) + (𝑎 + 2𝑥)+. . . +(𝑎 + (𝑛 − 2)𝑥) + (𝑎 + (𝑛 − 1)𝑥)
Summing this all together looks pretty tough. Instead, let’s ‘flip’ the terms around and put it side-by-side
with the original expression, in order to spot a pattern:
𝑆𝑛 = 𝑎 + (𝑎 + 𝑥) + (𝑎 + 2𝑥) + ⋯ + (𝑎 + (𝑛 − 2)𝑥) + (𝑎 + (𝑛 − 1)𝑥)
𝑆𝑛 = (𝑎 + (𝑛 − 1)𝑥) + (𝑎 + (𝑛 − 2)𝑥) + (𝑎 + (𝑛 − 2)𝑥)+. . . +(𝑎 + 𝑥) + 𝑎
‘Pair them up’ as shown above. You can see that each pair sums to: 2𝑎 + (𝑛 − 1)𝑥.
Since there are 𝑛 terms in the expression 𝑆𝑛 , then there are 𝑛 pairs in total as well.
If we sum the 2 expressions together, then, we get:
2𝑆𝑛 = 𝑛 × (2𝑎 + (𝑛 − 1)𝑥)
1
∴ 𝑆𝑛 = × 𝑛 × (2𝑎 + (𝑛 − 1)𝑥)
2
Remembering this formula would be horrendous, but fortunately we can see that the last term in the
sequence, call it 𝑙, was: 𝑙 = 𝑎 + (𝑛 − 1)𝑥.
∴ 2𝑎 + (𝑛 − 1)𝑥 = 𝑎 + (𝑎 + (𝑛 − 1)𝑥) = 𝑎 + 𝑙
…which is the first term + the last term of the series. And, substituting this into our expression for 𝑆𝑛 :
