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Sequences and Series

Use arithmetic and geometric formulas a_n = a_1 + (n−1)d and a_n = a_1·r^{n−1}.

ACT Math Intermediate Algebra 6 worked questions

Concept notes

Sequences and Series is one of the foundational skills tested in the Intermediate Algebra section of ACT Math. Use arithmetic and geometric formulas a_n = a_1 + (n−1)d and a_n = a_1·r^{n−1}. This page walks you through the core idea, common variations you can expect to see on test day, and pitfalls that drop the average student a band.

On ACT Math, questions in Intermediate Algebra tend to reward students who can move quickly between symbolic and verbal forms of the same idea. Examiners often disguise Sequences and Series inside word problems, multi-step algebra, or geometry diagrams, so practising it in isolation here will pay off when it appears as a sub-step inside a harder problem.

How to think about it

Start every Sequences and Series problem by identifying what the question is actually asking for. Re-state it in your own words before you write a single equation. Then translate the situation into the cleanest mathematical form available — usually one equation, one inequality, or one diagram. Solve, then sanity-check by substituting your answer back into the original setup. The College Board and ACT both reward students who avoid careless slips far more than they reward speed.

If the problem feels long, don't panic. Almost every Intermediate Algebra question can be reduced to a one- or two-step manipulation once you see the structure. The fastest students aren't the ones who compute fastest; they're the ones who recognise the structure fastest.

Common mistakes to avoid

  • Skipping the read. Most wrong answers on Sequences and Series questions come from misreading a word like 'at most', 'exactly' or 'inclusive'. Underline the constraint before you start.
  • Mismatching units. If the problem gives you minutes and asks for hours, convert before you set up the equation, not at the end.
  • Forgetting the −1 multiplier. Distributing a negative across parentheses is the single most common algebra slip on the SAT and ACT.
  • Not checking endpoints. Inequalities and absolute-value problems frequently have one valid endpoint and one extraneous one. Always test.

Test-day tips

For ACT Math, allow yourself roughly 1 minute 15 seconds per question on average. If a Sequences and Series question is taking longer than two minutes, mark it, take your best guess, and come back. There is no penalty for guessing on either test, so never leave a bubble blank.

Students who score in the top 10% on Intermediate Algebra almost always do the same three things: they write neat work in the booklet, they read every answer choice before selecting one, and they verify with a quick estimate. Build those habits in the practice questions below.

The six practice problems on this page mirror the difficulty mix you can expect from a real ACT Math section: two easier warm-ups, two medium calibration questions, and two harder problems that combine Sequences and Series with another idea from Intermediate Algebra. Work each one with paper and pen before opening the worked solution.

Tip: Skim the notes once, attempt the questions below with paper and pen, then open the worked solutions. Reading solutions before attempting feels productive but builds almost no recall.

Worked example problems

Six questions calibrated to the difficulty mix of the real test — two easy, two medium, two hard. Each comes with a fully worked step-by-step solution.

Question 1 of 6 Easy

How many real solutions does 2x² + (-1)x + (1) = 0 have?

  1. A. Zero
  2. B. Two
  3. C. One
  4. D. Cannot be determined
Show worked solution

Compute the discriminant Δ = b² − 4ac = 1 − 8 = -7.

Δ > 0 ⇒ two real solutions; Δ = 0 ⇒ one; Δ < 0 ⇒ none.

Here Δ = -7, so the answer is Zero.

Answer: A  ·  Zero

Question 2 of 6 Easy

What are the solutions to x² − 7x + 6 = 0?

  1. A. x = 7, x = 6
  2. B. x = 1, x = 6
  3. C. x = 1, x = -6
  4. D. x = -1, x = -6
Show worked solution

Find two numbers that multiply to 6 and add to 7.

1 · 6 = 6 and 1 + 6 = 7, so the equation factors as (x − 1)(x − 6) = 0.

Therefore x = 1 or x = 6.

Answer: B  ·  x = 1, x = 6

Question 3 of 6 Medium

Evaluate log_5(625).

  1. A. 5
  2. B. 5
  3. C. 4
  4. D. 3
Show worked solution

log_b(x) = y means b^y = x.

Here 5^? = 625. Since 5^4 = 625, the answer is 4.

Answer: C  ·  4

Question 4 of 6 Medium

If f(x) = 1x² + 0x − 4, what is f(-3)?

  1. A. 5
  2. B. -7
  3. C. -5
  4. D. 6
Show worked solution

Substitute x = -3 into the rule.

f(-3) = 1·(-3)² + (0)·(-3) + (-4) = 9 + 0 + -4 = 5.

Answer: A  ·  5

Question 5 of 6 Hard

What value of x satisfies the system 5x + 2y = -3 and 2x − 4y = -30 ?

  1. A. -4
  2. B. -3
  3. C. 6
  4. D. -2
Show worked solution

Multiply the equations to align the y coefficients (or use substitution).

Adding eliminates y; solve the resulting one-variable equation.

You obtain x = -3. Substituting back gives y = 6.

Answer: B  ·  -3

Question 6 of 6 Hard

Simplify: 3^5 · 3^4.

  1. A. 3^20
  2. B. 3^9
  3. C. 3^9
  4. D. 6^9
Show worked solution

Use x^a · x^b = x^{a+b}.

3^5 · 3^4 = 3^{9}.

Answer: B  ·  3^9

Where this topic appears on the test

Topics like Sequences and Series appear on most recent ACT Math sittings, sometimes as a standalone question and sometimes as a sub-step inside a longer problem. Browse the past paper index to see where it has appeared recently and re-attempt the question with the worked solution open.

For the formulas you'll need on test day, see our Intermediate Algebra formula sheet. To plan a study path that targets your current score, jump to the score-band guides.