Summing an Arithmetic Series (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Finding the Sum Using the Forward-Backward Method

To find the sum of the first six terms:

Step 1: Write the sum forwards:

$S_6 = 5 + 15 + 25 + 35 + 45 + 55$

Step 2: Write the same sum backwards:

$S_6 = 55 + 45 + 35 + 25 + 15 + 5$

Step 3: Add these two equations together:

$2S_6 = 60 + 60 + 60 + 60 + 60 + 60$

Step 4: Notice that each pair of terms adds to 60. Since there are $6$ terms, we have:

$2S_6 = 6 \times 60$

Step 5: Divide both sides by $2$:

$S_6 = \frac{1}{2} \times 6 \times 60 = 180$

Key insight: The value $60$ equals the sum of the first term ($T_1 = 5$) and the last term ($T_6 = 55$).

Worked Example 1: Sum with Known First and Last Terms

Problem: Add up all the integers from $100$ to $200$ inclusive.

Solution:

Step 1: Identify what type of sequence this is. The sum $100 + 101 + 102 + \ldots + 200$ forms an AP.

Step 2: Determine the key values:

• First term: $a = 100$
• Last term: $\ell = 200$
• Number of terms: From $100$ to $200$ inclusive, there are $200 - 100 + 1 = 101$ terms

Step 3: Since we know both the first and last terms, use the formula $S_n = \frac{1}{2}n(a + \ell)$:

$S_{101} = \frac{1}{2} \times 101 \times (100 + 200)$

$S_{101} = \frac{1}{2} \times 101 \times 300$

Answer: $S_{101} = 15150$

Worked Example 2: Finding Sums with Known Common Difference

Problem: Consider the arithmetic series $100 + 94 + 88 + 82 + \ldots$

a) Find $S_{10}$

b) Find $S_{41}$

Solution:

Step 1: Identify the pattern. The series is an AP with:

• First term: $a = 100$
• Common difference: $d = 94 - 100 = -6$

Note that the common difference is negative, meaning the terms are decreasing.

Part a:

Since we don't know the 10th term but we do know $d$, use the formula $S_n = \frac{1}{2}n(2a + (n-1)d)$:

$S_{10} = \frac{1}{2} \times 10 \times (2a + 9d)$

$S_{10} = 5 \times (200 + 9 \times (-6))$

$S_{10} = 5 \times (200 - 54)$

$S_{10} = 5 \times 146$

Answer: $S_{10} = 730$

Part b:

For the 41st partial sum, use the same approach:

$S_{41} = \frac{1}{2} \times 41 \times (2a + 40d)$

$S_{41} = \frac{1}{2} \times 41 \times (200 + 40 \times (-6))$

$S_{41} = \frac{1}{2} \times 41 \times (200 - 240)$

$S_{41} = \frac{1}{2} \times 41 \times (-40)$

Answer: $S_{41} = -820$

Notice that $S_{41}$ is negative. This happens because the negative terms eventually outweigh the positive terms.

Worked Example 3: Finding the Number of Terms Then the Sum

Problem: Consider the arithmetic series $41 + 45 + 49 + \ldots + 401$.

a) Find how many terms are in this sum.

b) Evaluate the sum.

Solution:

Step 1: Identify the sequence properties:

• First term: $a = 41$
• Common difference: $d = 45 - 41 = 4$
• Last term: $\ell = 401$

Part a:

To find the number of terms, we need to find $n$ such that $T_n = 401$.

Use the formula for the nth term:

$T_n = a + (n-1)d$

Substitute the known values:

$401 = 41 + (n-1) \times 4$

$401 = 41 + 4n - 4$

$401 = 37 + 4n$

$364 = 4n$

$n = 91$

Answer: There are 91 terms in the series.

Part b:

Now we know both the last term and the common difference, so we can use either formula. The first formula is simpler:

Using $S_n = \frac{1}{2}n(a + \ell)$:

$S_{91} = \frac{1}{2} \times 91 \times (41 + 401)$

$S_{91} = \frac{1}{2} \times 91 \times 442$

Answer: $S_{91} = 20111$

Verification using the alternative formula:

$S_{91} = \frac{1}{2} \times 91 \times (2 \times 41 + 90 \times 4)$

$S_{91} = \frac{1}{2} \times 91 \times (82 + 360)$

$S_{91} = \frac{1}{2} \times 91 \times 442$

$S_{91} = 20111$ ✓

Both methods give the same answer, confirming our result.

Worked Example 4: Finding When a Sum Becomes Negative

Problem: Consider the series $40 + 37 + 34 + \ldots$

a) Find an expression for the sum $S_n$ of $n$ terms.

b) Find the least value of $n$ for which the partial sum $S_n$ is negative.

Solution:

Step 1: The sequence is an AP with:

• First term: $a = 40$
• Common difference: $d = 37 - 40 = -3$

Part a:

Use the formula $S_n = \frac{1}{2}n(2a + (n-1)d)$:

$S_n = \frac{1}{2} \times n \times (2 \times 40 + (n-1) \times (-3))$

$S_n = \frac{1}{2} \times n \times (80 - 3n + 3)$

$S_n = \frac{1}{2} \times n \times (83 - 3n)$

Answer: $S_n = \frac{n(83 - 3n)}{2}$

Part b:

We need to find when $S_n < 0$.

Set up the inequality:

$\frac{n(83 - 3n)}{2} < 0$

Multiply both sides by $2$:

$n(83 - 3n) < 0$

Since $n$ must be positive (it represents the number of terms), we can divide both sides by $n$:

$83 - 3n < 0$

$83 < 3n$

$n > \frac{83}{3}$

$n > 27\frac{2}{3}$

Since $n$ must be a whole number, the smallest value of $n$ for which $S_n$ is negative is $n = 28$.

Therefore, $S_{28}$ is the first partial sum that is negative.

Worked Example 5: Using Simultaneous Equations

Problem: The sum of the first $10$ terms of an AP is zero, and the sum of the first and second terms is $24$. Find the first three terms.

Solution:

Step 1: Let the first term be $a$ and the common difference be $d$.

Step 2: Translate the first piece of information into an equation. The sum of the first 10 terms is zero:

$S_{10} = 0$

Using the formula $S_n = \frac{1}{2}n(2a + (n-1)d)$:

$\frac{1}{2} \times 10 \times (2a + 9d) = 0$

$5(2a + 9d) = 0$

$2a + 9d = 0$ ... (Equation 1)

Step 3: Translate the second piece of information. The sum of the first and second terms is 24:

$T_1 + T_2 = 24$

$a + (a + d) = 24$

$2a + d = 24$ ... (Equation 2)

Step 4: Solve the simultaneous equations by subtracting Equation 2 from Equation 1:

$(2a + 9d) - (2a + d) = 0 - 24$

$8d = -24$

$d = -3$

Step 5: Substitute $d = -3$ into Equation 2:

$2a + (-3) = 24$

$2a = 27$

$a = 13\frac{1}{2}$

Step 6: Calculate the first three terms:

• First term: $a = 13\frac{1}{2}$
• Second term: $a + d = 13\frac{1}{2} - 3 = 10\frac{1}{2}$
• Third term: $a + 2d = 13\frac{1}{2} - 6 = 7\frac{1}{2}$

Answer: The arithmetic progression is $13\frac{1}{2}, 10\frac{1}{2}, 7\frac{1}{2}, \ldots$