Arithmetic Series Revision Notes for Leaving Cert Mathematics

Arithmetic Series

What is an arithmetic series?

An arithmetic series is formed when we add together the terms of an arithmetic sequence. This is different from the sequence itself, which is just a list of numbers.

For example:

Arithmetic sequence: 1, 3, 5, 7, ... (just the numbers)
Arithmetic series: 1 + 3 + 5 + 7 + ... (adding the numbers together)

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The key difference is that a sequence shows the pattern of numbers, while a series shows us adding those numbers together to find their total.

Notation and key terms

We use specific notation to work with arithmetic series:

S_n represents the sum of the first n terms of a series
a is the first term of the series
d is the common difference between consecutive terms
T_n represents the nth individual term

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Understanding the notation:

For example:

S₁ = sum of first 1 term = T₁
S₂ = T₁ + T₂
S₃ = T₁ + T₂ + T₃

The sum formula

The sum formula for an arithmetic series allows us to calculate the total of the first n terms without having to add each term individually.

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The Sum Formula for Arithmetic Series:

$S_n = \frac{n}{2}[2a + (n-1)d]$

Where:

n = number of terms we want to add
a = first term
d = common difference

This formula works because it calculates the average of the first and last terms, then multiplies by the number of terms.

Worked example 1: Finding the sum

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Worked Example: Finding the Sum

Question: Find S_n and hence S₂₀ of the series 5 + 8 + 11 + 14 + ...

Solution:

Step 1: Identify the values

First term: a = 5
Common difference: d = 3 (since 8 - 5 = 3)

Step 2: Apply the formula $S_n = \frac{n}{2}[2a + (n-1)d]$

Step 3: Substitute our values $S_n = \frac{n}{2}[2(5) + (n-1)(3)]$ $S_n = \frac{n}{2}[10 + 3n - 3]$ $S_n = \frac{n}{2}[7 + 3n]$ $S_n = \frac{n(3n + 7)}{2}$

Step 4: Find S₂₀ by substituting n = 20 $S_{20} = \frac{20}{2}[60 + 7] = 10(67) = 670$

Worked example 2: Finding the number of terms

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Worked Example: Finding the Number of Terms

Question: Given the arithmetic series 5 + 7 + 9 + ..., if S_n = 192, find the value of n.

Solution:

Step 1: Identify the values

a = 5, d = 2

Step 2: Set up the equation using the sum formula $\frac{n}{2}[2a + (n-1)d] = 192$

Step 3: Substitute the known values $\frac{n}{2}[10 + (n-1)(2)] = 192$ $\frac{n}{2}[10 + 2n - 2] = 192$ $\frac{n}{2}[8 + 2n] = 192$ $\frac{n(2n + 8)}{2} = 192$

Step 4: Solve the equation Multiply both sides by 2: $n(2n + 8) = 384$ $2n² + 8n = 384$ $n² + 4n - 192 = 0$

Step 5: Factor and solve $(n - 12)(n + 16) = 0$

Therefore n = 12 (we disregard the negative answer -16)

So S₁₂ = 192

Finding individual terms from sums

There's an important relationship between consecutive sums and individual terms that allows us to find any term in the series.

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Key Relationship:

$T_n = S_n - S_{n-1}$

This means the nth term equals the sum of n terms minus the sum of (n-1) terms.

Worked example 3: Finding terms from a sum formula

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Worked Example: Finding Terms from a Sum Formula

Question: In an arithmetic series, S_n = n² + 2n. Find S₁, S₂ and S₃, and hence write down T₁, T₂ and T₃.

Solution:

Step 1: Calculate the sums

S₁ = 1² + 2(1) = 3, so T₁ = 3
S₂ = 2² + 2(2) = 8, so T₁ + T₂ = 8, therefore T₂ = 5
S₃ = 3² + 2(3) = 15, so T₁ + T₂ + T₃ = 15, therefore T₃ = 7

Step 2: Identify the terms The first three terms are 3, 5, 7.

Exam tips

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Exam Strategy Tips:

Always identify a and d first before using the formula
When finding n, you'll often get a quadratic equation - always check your answer makes sense (n must be positive)
Remember that T_n = S_n - S_(n-1) is useful for finding individual terms
Double-check your arithmetic, especially when expanding brackets

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Key Points to Remember:

Arithmetic series = adding the terms of an arithmetic sequence together
Key formula: $S_n = \frac{n}{2}[2a + (n-1)d]$
Individual terms: $T_n = S_n - S_{n-1}$
Always identify a (first term) and d (common difference) before applying formulas
When solving for n, disregard negative solutions as they don't make physical sense

Arithmetic Series (Leaving Cert Mathematics): Revision Notes