Arithmetic sequences and series; sigma notation practice questions.

The third term of an arithmetic sequence is 11 and the seventh term is 27. Determine the common difference.

• 2
• 4
• 5
• 16

Answer: 4

Between the 3rd and 7th terms there are 7 − 3 = 4 steps, so 4d = 27 − 11 = 16 and d = 4. Leaving the answer as 16 forgets to divide by the number of steps.

An arithmetic sequence has first term 40 and common difference −3. Determine which term is the first one to be negative.

• the 13th term
• the 14th term
• the 15th term
• the 16th term

Answer: the 15th term

un = 40 − 3(n − 1) < 0 gives n > 43/3 ≈ 14.3, so n = 15. Check: u14 = 40 − 39 = 1 (still positive) and u15 = 40 − 42 = −2.

Evaluate the sum Σ (2r + 1) for r = 1 to 5.

• 25
• 30
• 45
• 35

Answer: 35

The terms are 3, 5, 7, 9, 11, which sum to 35. Forgetting the +1 in each term gives the distractor 30.

A theatre has 18 seats in the first row, and each row after the first has 4 more seats than the row before it. Calculate the number of seats in row 12.

• 62
• 66
• 58
• 48

Answer: 62

u12 = 18 + (12 − 1) × 4 = 18 + 44 = 62. Using 12 × 4 instead of 11 × 4 gives the classic distractor 66.

The sum of the first n terms of an arithmetic series is given by Sn = 2n² + 3n. The first term of the series is

• 2
• 3
• 5
• 7

Answer: 5

The first term equals S1 = 2(1)² + 3(1) = 5. Reading off a coefficient of the formula instead of substituting n = 1 gives the distractors.

An arithmetic series has first term 7 and common difference 5. Determine the smallest value of n for which the sum of the first n terms exceeds 400.

• 11
• 12
• 13
• 14

Answer: 12

Sn = n/2 (2 × 7 + 5(n − 1)) = n(5n + 9)/2. S11 = 352 and S12 = 414, so n = 12 is the first sum above 400.