Chapter 2 · Sequences · Concept 1
2-6 Sum of Arithmetic Sequence (3)
A lake loses the same depth of water every day. Its total loss keeps growing — but a business that loses a little less money each month eventually turns a corner. Sequences that fall have a moment where adding one more term starts making things worse, and finding that moment is a question in itself.
Guiding question
When does adding one more term stop helping?
Learning outcome. Find the sum of an arithmetic sequence, and determine the number of terms in it.
Same two formulas, one new idea
Sₙ = ½n(a + l)
Sₙ = ½n{2a + (n − 1)d}
To find the biggest possible total of a sequence that is going down, add up the positive terms — and stop.
11, 8, 5, 2, −1, −4, …
The total is biggest at 2. Everything after it is taking away.
The top bars are the running total. The bottom bars are the terms themselves. Drag along and watch what happens to the total at the exact moment the terms go red.
the running total Sₙ
the terms aₙ themselves
term 1
65
total so far S1
65
The total climbs while the terms are green and falls once they turn red. So the biggest total is reached at term 17 — the last positive one — and it is 561.
−4n + 69 > 0 gives n < 16.25, and the biggest whole number below that is 17.
This is the mistake the textbook warns about by name. Both boxes below are honest algebra, carefully done. Only one of them answers the question that was asked.
the right question
aₙ > 0
“Up to which term are we still adding something?” Once a term is negative, including it makes the total smaller.
−4n + 69 > 0 → n < 17.25
n = 17 ✓
the wrong question
Sₙ > 0
“Up to which term is the total still positive?” That is a different thing — the total stays positive long after it has started shrinking.
½n(134 − 4n) > 0 → n < 33.5
n = 33 ✗
Why the totals stay positive so much longer.
After term 17 the total starts shrinking — but it starts from 561, so it takes a long time to eat all of that away. It is still positive at term 34. Being positive and being biggest are simply not the same thing, and the question asks for biggest.
1. First term 4, common difference 3. How many terms add up to 50?
- 1
Put what you know into the sum formula, leaving n as a letter.
50 = ½n{2 × 4 + (n − 1) × 3}This time n is the unknown, not the answer you were given. Everything else goes in.
3. First term 65, common difference −4. How many terms give the biggest sum, and what is it?
- 1
Find the general term first.
aₙ = 65 + (n − 1)(−4) = −4n + 69You cannot ask when a term turns negative until you can write the term.
Work with a partner. For question 3, one of you finds where the terms turn negative and the other does the sum. Then swap and check each other.
The 15th term of an arithmetic sequence is 33 and the 50th term is 103. What is the sum of the first 79 terms?
If the water level in a lake drops by a fixed amount each day, how do arithmetic sequences let you predict when the water will run out — and what does the 'maximum sum' idea mean in that story?
What is the term, and what is the running total?
Solving Sₙ > 0 for the maximum sum.
The question is where the TERMS turn negative, so solve aₙ > 0. The total stays positive far longer than it stays growing — those are different n.
Keeping the impossible root.
A quadratic in n gives two answers, and one is usually negative or a fraction. Say the sentence: "n is a whole positive number, so n = 5."
Forgetting to answer both halves.
"Find n for the maximum sum, and find that sum" is two marks. Giving only n throws one away.
Dropping the half when you clear the fraction.
50 = ½n(3n + 5) becomes 100 = n(3n + 5). Double BOTH sides, not just the one with the fraction in it.
Rounding the boundary the wrong way.
n < 17.25 for a POSITIVE term means the last good one is 17. Compare with lesson 2-3, where n > 124.75 meant the first good one was 125. Read which direction the inequality points.
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