Arithmetic Series
This topic covers arithmetic sequences, where numbers increase or decrease by a constant amount. For ESAT, you must be able to find any term in a sequence and calculate the sum of a series efficiently without a calculator.
Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- An arithmetic sequence is defined by its first term (a) and a constant common difference (d) between consecutive terms.
- The formula for the nth term, un = a + (n-1)d, is a linear relationship in n. Recognising this can help solve problems without direct formula use.
- The sum of an arithmetic series is the number of terms multiplied by the average of the first and last terms.
- The sum of the first n positive integers (1 + 2 +⋯ + n) is a special case of an arithmetic series with the formula Sn = n(n+1)/2.
- Adding two arithmetic sequences term-by-term creates a new arithmetic sequence. Its first term is the sum of the original first terms, and its common difference is the sum of the original common differences.
Formulae
un = a + (n-1)d To find the value of a specific term (the nth term) in the sequence, given the first term 'a' and common difference 'd'.
Sn = n/2 × (2a + (n-1)d) To find the sum of the first 'n' terms when you know the first term 'a', the common difference 'd', and the number of terms 'n'.
Sn = n/2 × (a + un) A faster way to find the sum of the first 'n' terms if you already know the first term 'a' and the last term 'un'.
Sn = n(n+1)/2 A specific shortcut to find the sum of the first n positive integers (1, 2, 3, ⋯, n).
Definitions
- Arithmetic Sequence / Progression
- A sequence of numbers where each term after the first is found by adding a constant, the common difference, to the previous one.
- Common Difference (d)
- The fixed amount added to each term to get the next term in an arithmetic sequence. It can be positive, negative, or zero.
- Arithmetic Series
- The result of adding up the terms of an arithmetic sequence.
Worked example
The fourth term of an arithmetic sequence is 11 and the tenth term is 29. What is the sum of the first 15 terms of this sequence?
- 1
Set up simultaneous equations using the nth term formula, un = a + (n-1)d.
- 2
For the 4th term:
u4 = a + (4-1)d = a + 3d = 11 - 3
For the 10th term:
u10 = a + (10-1)d = a + 9d = 29 - 4
Subtract the first equation from the second:
(a + 9d) - (a + 3d) = 29 - 11, which simplifies to 6d = 18, so d = 3.
- 5 Substitute d=3 into the first equation:a + 3(3) = 11, which gives a + 9 = 11, so a = 2
- 6
Now find the sum of the first 15 terms using Sn = n/2 × (2a + (n-1)d) with n=15, a=2, d=3.
- 7 S15 = 15/2 × (2(2) + (15-1)3) = 15/2 × (4 + 14*3) = 15/2 × (4 + 42) = 15/2 × (46)
- 8
Calculate the final sum:
15 × (46/2) = 15 × 23 = 345
Answer: 345
Common mistakes
- ×Simple calculation mistakes under pressure are very frequent, especially when dealing with negative values for 'a' or 'd'. Always double-check your arithmetic.
- ×Using 'n' instead of '(n-1)' in the term and sum formulas. This is a classic off-by-one error that leads to an incorrect answer by a factor of 'd'.
- ×Confusing the formula for the nth term (un) with the formula for the sum of n terms (Sn). Be clear whether the question asks for a single term's value or a cumulative total.
No-calculator tips
- ✓When calculating a sum Sn = n/2 × (first + last), always check if 'n' or '(first + last)' is even. Divide the even number by 2 first to simplify the multiplication.
- ✓To calculate a product like 15 × 23, break it down: 15 × 20 = 300, and 15 × 3 = 45. Then add them: 300 + 45 = 345.
- ✓For the sum of natural numbers, Sn = n(n+1)/2, one of n or n+1 must be even. Perform the division by 2 on the even number before multiplying to keep the numbers smaller.