An arithmetic sequence has a first term of 5 and a common difference of 3. What is the 8th term?
A24
B26
C29
D21
a_8 = 5 + (8−1)(3) = 5 + 21 = 26. The formula uses (n−1) because you apply the common difference one fewer time than the term number — you start at a_1 before adding any d. The most tempting wrong answer is option 2 (29), which comes from computing 5 + 8(3) and adding d eight times instead of seven. A quick check: a_2 = 5+3 = 8, a_3 = 11, a_4 = 14… a_8 = 26.
Question 2 Multiple Choice
A sequence starts at 2 with a common difference of 4. Which expression correctly gives the sum of the first 10 terms?
A10 × 2 + 10 × 4
B(10/2)(2 × 2 + 9 × 4)
C2 + (10−1) × 4
D10 × (2 + 10 × 4) / 2
The sum formula is S_n = (n/2)(2a_1 + (n−1)d). With n=10, a_1=2, d=4: S_10 = (10/2)(4 + 36) = 5 × 40 = 200. Option 2 gives the 10th *term* a_10 = 2 + 9×4 = 38, not the sum — a classic confusion between the nth term formula and the sum formula. Options 0 and 3 are not valid formulas for either quantity.
Question 3 True / False
An arithmetic sequence and a geometric sequence can both exhibit a regular, predictable pattern, so you must check whether the pattern is additive or multiplicative to distinguish them.
TTrue
FFalse
Answer: True
Both sequence types are regular, but their regularity is structurally different. Arithmetic sequences add a constant difference each step (5, 8, 11, 14…). Geometric sequences multiply by a constant ratio each step (2, 6, 18, 54…). A student seeing a 'pattern' cannot assume arithmetic — they must check: are the differences constant? Or are the ratios constant? The formulas, sum formulas, and long-run behavior are completely different for the two types.
Question 4 True / False
In the formula a_n = a_1 + (n−1)d, the (n−1) factor appears because the first term does not count as an application of the common difference.
TTrue
FFalse
Answer: True
This is the conceptual reason for the (n−1). The first term a_1 exists before any common difference has been added — it is the starting value. To reach the nth term, d is applied exactly (n−1) times: a_2 = a_1 + d (one application), a_3 = a_1 + 2d (two applications), a_n = a_1 + (n−1)d. The common error is writing a_1 + nd, which overcounts by one application and produces a value that is d too large.
Question 5 Short Answer
Explain Gauss's trick for summing an arithmetic series. Why does pairing the first and last terms work?
Think about your answer, then reveal below.
Model answer: Write the sum forwards and backwards, then add term-by-term. Each of the n pairs sums to (a_1 + a_n) because gaining one step from the front is exactly offset by losing one step from the back. So 2S_n = n(a_1 + a_n), giving S_n = n(a_1 + a_n)/2. For 1+2+…+100: pairing 1+100 = 101, 2+99 = 101, etc. gives 50 pairs × 101 = 5050.
The trick works because in an arithmetic sequence, the common difference cancels when you pair symmetrically from opposite ends. Each step forward from a_1 adds d; each step backward from a_n subtracts d. The sum of any symmetric pair is therefore constant — always (a_1 + a_n) — regardless of which pair you pick. This elegant symmetry is what makes the sum formula so clean.