IB Mathematics: Analysis and Approaches Quiz: Arithmetic Sequences And Series
20 questions · exam conditions
0:00
Arithmetic Sequences And SeriesQuestion 1 of 20

The first term of arithmetic sequence A is 5 and its common difference is 2. The first term of arithmetic sequence B is 2 and its common difference is 3.

Find the value of nn for which the nn-th terms of the two sequences are equal.

3
4
5
7
← Back to quizzes

IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Arithmetic Sequences And Series

Practice Arithmetic Sequences And Series in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Arithmetic Sequences And Series, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The first term of arithmetic sequence A is 5 and its common difference is 2. The first term of arithmetic sequence B is 2 and its common difference is 3.

Find the value of nn for which the nn-th terms of the two sequences are equal.

  1. 3
  2. 4 (correct answer)
  3. 5
  4. 7
Explanation: Let AnA_n be the nn-th term of sequence A, and BnB_n be the nn-th term of sequence B. The formula for the nn-th term is un=u1+(n−1)du_n = u_1 + (n-1)d. For sequence A: An=5+(n−1)2=5+2n−2=2n+3A_n = 5 + (n-1)2 = 5 + 2n - 2 = 2n + 3. For sequence B: Bn=2+(n−1)3=2+3n−3=3n−1B_n = 2 + (n-1)3 = 2 + 3n - 3 = 3n - 1. We want to find nn such that An=BnA_n = B_n. 2n+3=3n−12n + 3 = 3n - 1 Subtract 2n2n from both sides: 3=n−13 = n - 1. Add 1 to both sides: 4=n4 = n. So, the 4th terms are equal. Distractor Explanations: A. 3: This results from using ndnd instead of (n−1)d(n-1)d: 5+2n=2+3n⇒n=35+2n = 2+3n \Rightarrow n=3. C. 5: A possible calculation error. D. 7: This is the value of nn for which the sums of the two sequences are equal, SA,n=SB,nS_{A,n} = S_{B,n}. SA,n=n2(10+(n−1)2)=n(n+4)S_{A,n} = \frac{n}{2}(10+(n-1)2) = n(n+4). SB,n=n2(4+(n−1)3)=n2(3n+1)S_{B,n} = \frac{n}{2}(4+(n-1)3) = \frac{n}{2}(3n+1). Setting them equal: n2+4n=3n2+n2⇒2n2+8n=3n2+n⇒n2−7n=0⇒n(n−7)=0n^2+4n = \frac{3n^2+n}{2} \Rightarrow 2n^2+8n = 3n^2+n \Rightarrow n^2-7n=0 \Rightarrow n(n-7)=0. Since n>0n>0, n=7n=7.

Question 2

A company's profit was $120,000 in its first year. Each year, the profit increased by a constant amount of $8,000.

In which year will the company's total profit since it started first exceed $2,000,000?

  1. 12th
  2. 237th
  3. 17th
  4. 13th (correct answer)
Explanation: This question tests arithmetic sequences and series - when you see constant increases over time, you're dealing with an arithmetic progression where you need to find when the cumulative sum reaches a target. The company starts with $120,000 profit in year 1, then gains $8,000 each subsequent year. This creates the sequence: $120,000, $128,000, $136,000, etc. The total profit after n years is the sum of this arithmetic series: $Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d] wherewherea = 120,000andandd = 8,000$. Substituting: S_n = \frac{n}{2}[240,000 + 8,000(n-1)] = \frac{n}{2}[232,000 + 8,000n] = 4,000n(58 + n) We need Sn>2,000,000S_n > 2,000,000, so: 4,000n(58 + n) > 2,000,000 n(58 + n) > 500 n^2 + 58n - 500 > 0 Using the quadratic formula: n=−58+582+4(500)2≈7.6n = \frac{-58 + \sqrt{58^2 + 4(500)}}{2} \approx 7.6 Since we need the first year to exceed $2,000,000, we test: $S_{12} = 4,000(12)(70) = 3,360,000andandS_{13} = 4,000(13)(71) = 3,692,000.Bothexceed$2,000,000,butcheckingbackwards:$S12=3,360,000>2,000,000. Both exceed $2,000,000, but checking backwards: $S_{12} = 3,360,000 > 2,000,000, so year 13 is correct. Choice A (12th) gives exactly when it's exceeded, but the question asks "in which year" it first exceeds. Choice B (237th) likely comes from incorrectly using annual profit instead of cumulative. Choice C (17th) may result from calculation errors in the quadratic solution. Always distinguish between "when does it first exceed" versus "what's the minimum value" - they often differ by one unit in sequence problems.

Question 3

The first term of arithmetic sequence A is 5 and its common difference is 2. The first term of arithmetic sequence B is 2 and its common difference is 3.

Find the value of nn for which the nn-th terms of the two sequences are equal.

  1. 3
  2. 4 (correct answer)
  3. 5
  4. 7
Explanation: Let AnA_n be the nn-th term of sequence A, and BnB_n be the nn-th term of sequence B. The formula for the nn-th term is un=u1+(n−1)du_n = u_1 + (n-1)d. For sequence A: An=5+(n−1)2=5+2n−2=2n+3A_n = 5 + (n-1)2 = 5 + 2n - 2 = 2n + 3. For sequence B: Bn=2+(n−1)3=2+3n−3=3n−1B_n = 2 + (n-1)3 = 2 + 3n - 3 = 3n - 1. We want to find nn such that An=BnA_n = B_n. 2n+3=3n−12n + 3 = 3n - 1 Subtract 2n2n from both sides: 3=n−13 = n - 1. Add 1 to both sides: 4=n4 = n. So, the 4th terms are equal. Distractor Explanations: A. 3: This results from using ndnd instead of (n−1)d(n-1)d: 5+2n=2+3n⇒n=35+2n = 2+3n \Rightarrow n=3. C. 5: A possible calculation error. D. 7: This is the value of nn for which the sums of the two sequences are equal, SA,n=SB,nS_{A,n} = S_{B,n}. SA,n=n2(10+(n−1)2)=n(n+4)S_{A,n} = \frac{n}{2}(10+(n-1)2) = n(n+4). SB,n=n2(4+(n−1)3)=n2(3n+1)S_{B,n} = \frac{n}{2}(4+(n-1)3) = \frac{n}{2}(3n+1). Setting them equal: n2+4n=3n2+n2⇒2n2+8n=3n2+n⇒n2−7n=0⇒n(n−7)=0n^2+4n = \frac{3n^2+n}{2} \Rightarrow 2n^2+8n = 3n^2+n \Rightarrow n^2-7n=0 \Rightarrow n(n-7)=0. Since n>0n>0, n=7n=7.

Question 4

The sum of the first 10 terms of an arithmetic series is 120. If the first term is 3, what is the value of the sixth term, u6u_6?

  1. -7
  2. 2
  3. 13 (correct answer)
  4. 15
Explanation: First, find the common difference, dd, using the sum formula Sn=n2(2u1+(n−1)d)S_n = \frac{n}{2}(2u_1 + (n-1)d). We have S10=120S_{10}=120, u1=3u_1=3, and n=10n=10. 120=102(2(3)+(10−1)d)120 = \frac{10}{2}(2(3) + (10-1)d) 120=5(6+9d)120 = 5(6 + 9d) 24=6+9d24 = 6 + 9d 18=9d⇒d=218 = 9d \Rightarrow d=2. Now, find the sixth term using the formula un=u1+(n−1)du_n = u_1 + (n-1)d. u6=3+(6−1)(2)=3+5(2)=3+10=13u_6 = 3 + (6-1)(2) = 3 + 5(2) = 3 + 10 = 13. Distractor Explanations: A. -7: This is the result of a sign error when finding dd. If 9d=−18⇒d=−29d = -18 \Rightarrow d=-2, then u6=3+5(−2)=−7u_6 = 3 + 5(-2) = -7. B. 2: This is the value of the common difference dd, not the sixth term. D. 15: This is a common off-by-one error, using ndnd instead of (n−1)d(n-1)d to find the term: u6=u1+6d=3+6(2)=15u_6 = u_1 + 6d = 3 + 6(2) = 15.

Question 5

A concert hall has 25 rows of seats. The first row has 20 seats, and each subsequent row has 2 more seats than the row in front of it. The price for a seat in the first 10 rows is $50, and the price for a seat in any of the remaining rows is $40.

Find the total revenue from a full house.

  1. $44,000
  2. $46,900 (correct answer)
  3. $49,500
  4. $52,100
Explanation: This is a two-part problem. First, find the number of seats in each section, then calculate the revenue. The number of seats forms an arithmetic sequence with u1=20u_1=20 and d=2d=2. Part 1: Seats and revenue for the first 10 rows (n=10n=10). Number of seats: S10=102(2(20)+(10−1)2)=5(40+18)=5(58)=290S_{10} = \frac{10}{2}(2(20) + (10-1)2) = 5(40 + 18) = 5(58) = 290 seats. Revenue: 290 \times \50 = $14,500.Part2:Seatsandrevenuefortheremainingrows(rows11to25,whichis. Part 2: Seats and revenue for the remaining rows (rows 11 to 25, which is 25-11+1=15rows).Totalseatsinall25rows:rows). Total seats in all 25 rows:S_{25} = \frac{25}{2}(2(20) + (25-1)2) = \frac{25}{2}(40 + 48) = \frac{25}{2}(88) = 25 \times 44 = 1100seats.Seatsinrows11−25:seats. Seats in rows 11-25:S_{25} - S_{10} = 1100 - 290 = 810seats.Revenue:seats. Revenue:810 \times $40 = $32,400.Part3:Totalrevenue.TotalRevenue=. Part 3: Total revenue. Total Revenue = $14,500 + $32,400 = $46,900$. Distractor Explanations: A. 44,000:Thisisthetotalnumberofseats(1100)multipliedbythecheaperprice(44,000: This is the total number of seats (1100) multiplied by the cheaper price (40). C. $49,500: This result may come from calculating the total seats (1100) and multiplying by an average price, such as $45. D. $52,100: This is the result of swapping the prices for the sections: 290 \times \40 + 810 \times $50 = $11,600 + $40,500 = $52,100$.

Question 6

A concert hall has 25 rows of seats. The first row has 20 seats, and each subsequent row has 2 more seats than the row in front of it. The price for a seat in the first 10 rows is $50, and the price for a seat in any of the remaining rows is $40.

Find the total revenue from a full house.

  1. $44,000
  2. $46,900 (correct answer)
  3. $49,500
  4. $52,100
Explanation: This is a two-part problem. First, find the number of seats in each section, then calculate the revenue. The number of seats forms an arithmetic sequence with u1=20u_1=20 and d=2d=2. Part 1: Seats and revenue for the first 10 rows (n=10n=10). Number of seats: S10=102(2(20)+(10−1)2)=5(40+18)=5(58)=290S_{10} = \frac{10}{2}(2(20) + (10-1)2) = 5(40 + 18) = 5(58) = 290 seats. Revenue: 290 \times \50 = $14,500.Part2:Seatsandrevenuefortheremainingrows(rows11to25,whichis. Part 2: Seats and revenue for the remaining rows (rows 11 to 25, which is 25-11+1=15rows).Totalseatsinall25rows:rows). Total seats in all 25 rows:S_{25} = \frac{25}{2}(2(20) + (25-1)2) = \frac{25}{2}(40 + 48) = \frac{25}{2}(88) = 25 \times 44 = 1100seats.Seatsinrows11−25:seats. Seats in rows 11-25:S_{25} - S_{10} = 1100 - 290 = 810seats.Revenue:seats. Revenue:810 \times $40 = $32,400.Part3:Totalrevenue.TotalRevenue=. Part 3: Total revenue. Total Revenue = $14,500 + $32,400 = $46,900$. Distractor Explanations: A. 44,000:Thisisthetotalnumberofseats(1100)multipliedbythecheaperprice(44,000: This is the total number of seats (1100) multiplied by the cheaper price (40). C. $49,500: This result may come from calculating the total seats (1100) and multiplying by an average price, such as $45. D. $52,100: This is the result of swapping the prices for the sections: 290 \times \40 + 810 \times $50 = $11,600 + $40,500 = $52,100$.

Question 7

An arithmetic sequence has first term u1=5u_1=5 and common difference d=3d=3. Find the sum of the terms from the 11th to the 20th, inclusive.

  1. 450
  2. 485 (correct answer)
  3. 517
  4. 670
Explanation: The sum required is u11+u12+⋯+u20u_{11} + u_{12} + \dots + u_{20}. This can be calculated as the sum of the first 20 terms minus the sum of the first 10 terms (S20−S10S_{20} - S_{10}). First, calculate S20S_{20}: S20=202(2(5)+(20−1)(3))=10(10+19×3)=10(10+57)=670S_{20} = \frac{20}{2}(2(5) + (20-1)(3)) = 10(10 + 19 \times 3) = 10(10 + 57) = 670. Next, calculate S10S_{10}: S10=102(2(5)+(10−1)(3))=5(10+9×3)=5(10+27)=5(37)=185S_{10} = \frac{10}{2}(2(5) + (10-1)(3)) = 5(10 + 9 \times 3) = 5(10 + 27) = 5(37) = 185. The required sum is S20−S10=670−185=485S_{20} - S_{10} = 670 - 185 = 485. Alternatively, we can treat this as a new arithmetic series with n=10n=10 terms. The first term is a1=u11=5+(11−1)(3)=35a_1 = u_{11} = 5 + (11-1)(3) = 35. The last term is a10=u20=5+(20−1)(3)=62a_{10} = u_{20} = 5 + (20-1)(3) = 62. The sum is 102(35+62)=5(97)=485\frac{10}{2}(35 + 62) = 5(97) = 485. Distractor Explanations: A. 450: This is the result of an off-by-one error in the range, calculating S20−S11S_{20} - S_{11}. C. 517: This is the result of another common off-by-one error, calculating S20−S9S_{20} - S_9. D. 670: This is the sum of the first 20 terms, S20S_{20}, without subtracting the first 10.

Question 8

A company's profit was $120,000 in its first year. Each year, the profit increased by a constant amount of $8,000.

In which year will the company's total profit since it started first exceed $2,000,000?

  1. 12th
  2. 237th
  3. 17th
  4. 13th (correct answer)
Explanation: This question tests arithmetic sequences and series - when you see constant increases over time, you're dealing with an arithmetic progression where you need to find when the cumulative sum reaches a target. The company starts with $120,000 profit in year 1, then gains $8,000 each subsequent year. This creates the sequence: $120,000, $128,000, $136,000, etc. The total profit after n years is the sum of this arithmetic series: $Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d] wherewherea = 120,000andandd = 8,000$. Substituting: S_n = \frac{n}{2}[240,000 + 8,000(n-1)] = \frac{n}{2}[232,000 + 8,000n] = 4,000n(58 + n) We need Sn>2,000,000S_n > 2,000,000, so: 4,000n(58 + n) > 2,000,000 n(58 + n) > 500 n^2 + 58n - 500 > 0 Using the quadratic formula: n=−58+582+4(500)2≈7.6n = \frac{-58 + \sqrt{58^2 + 4(500)}}{2} \approx 7.6 Since we need the first year to exceed $2,000,000, we test: $S_{12} = 4,000(12)(70) = 3,360,000andandS_{13} = 4,000(13)(71) = 3,692,000.Bothexceed$2,000,000,butcheckingbackwards:$S12=3,360,000>2,000,000. Both exceed $2,000,000, but checking backwards: $S_{12} = 3,360,000 > 2,000,000, so year 13 is correct. Choice A (12th) gives exactly when it's exceeded, but the question asks "in which year" it first exceeds. Choice B (237th) likely comes from incorrectly using annual profit instead of cumulative. Choice C (17th) may result from calculation errors in the quadratic solution. Always distinguish between "when does it first exceed" versus "what's the minimum value" - they often differ by one unit in sequence problems.

Question 9

Four arithmetic means are inserted between -3 and 17. Find the sum of these four means.

  1. 4
  2. 7
  3. 28 (correct answer)
  4. 42
Explanation: When four arithmetic means are inserted between -3 and 17, we get a new arithmetic sequence with a total of 4+2=64+2=6 terms. Let this sequence be unu_n. The first term is u1=−3u_1 = -3 and the sixth term is u6=17u_6 = 17. We can find the common difference dd using the formula un=u1+(n−1)du_n = u_1 + (n-1)d. u6=u1+(6−1)d⇒17=−3+5du_6 = u_1 + (6-1)d \Rightarrow 17 = -3 + 5d. 20=5d⇒d=420 = 5d \Rightarrow d=4. The sequence is -3, 1, 5, 9, 13, 17. The four arithmetic means are 1, 5, 9, and 13. Their sum is 1+5+9+13=281+5+9+13 = 28. Alternatively, the sum of an arithmetic sequence is the number of terms times the average of the first and last term. The sum of the two endpoints is −3+17=14-3+17=14. The sum of all 6 terms is S6=62(−3+17)=3(14)=42S_6 = \frac{6}{2}(-3+17) = 3(14)=42. The sum of the means is this total sum minus the endpoints: 42−(−3)−17=2842 - (-3) - 17 = 28. Distractor Explanations: A. 4: This is the value of the common difference dd, not the sum of the means. B. 7: This is the average of the first and last term, (−3+17)/2=7(-3+17)/2 = 7. The sum is 4 times this average. D. 42: This is the sum of all six terms in the sequence, including the endpoints -3 and 17.

Question 10

In an arithmetic sequence, the sum of the second and sixth terms is 32. The product of the first and fourth terms is 64. Find the common difference, dd.

  1. 2
  2. 4 (correct answer)
  3. 8
  4. 16
Explanation: Let the first term be u1u_1 and the common difference be dd. From the given information, we can write two equations:
  1. u2+u6=32u_2 + u_6 = 32 (u1+d)+(u1+5d)=32(u_1 + d) + (u_1 + 5d) = 32 2u1+6d=322u_1 + 6d = 32 Dividing by 2, we get u1+3d=16u_1 + 3d = 16. Note that the left side is the formula for the fourth term, u4u_4. So, u4=16u_4 = 16.
  2. u1⋅u4=64u_1 \cdot u_4 = 64 Substitute u4=16u_4 = 16 into the second equation: u1⋅16=64u_1 \cdot 16 = 64 u1=4u_1 = 4. Now use the equation u1+3d=16u_1 + 3d = 16 to find dd: 4+3d=164 + 3d = 16 3d=123d = 12 d=4d = 4.
Distractor Explanations: A. 2: This value for u1u_1 is obtained if one incorrectly simplifies 2u1+6d=322u_1+6d=32 to u1+3d=32u_1+3d=32, then substitutes into u1(u1+3d)=64u_1(u_1+3d)=64 to get u1(32)=64⇒u1=2u_1(32)=64 \Rightarrow u_1=2. C. 8: This is the value of the second term, u2=u1+d=4+4=8u_2 = u_1+d = 4+4=8. D. 16: This is the value of the fourth term, u4u_4.

Question 11

The sum of the first nn terms of an arithmetic sequence is given by the formula Sn=An2+BnS_n = An^2 + Bn. Which expression represents the common difference of the sequence?

  1. A
  2. B
  3. A + B
  4. 2A (correct answer)
Explanation: To find the common difference, we can find the first two terms and subtract them. The first term is u1=S1u_1 = S_1. u1=A(1)2+B(1)=A+Bu_1 = A(1)^2 + B(1) = A + B. The sum of the first two terms is S2=A(2)2+B(2)=4A+2BS_2 = A(2)^2 + B(2) = 4A + 2B. The second term is u2=S2−S1u_2 = S_2 - S_1. u2=(4A+2B)−(A+B)=3A+Bu_2 = (4A + 2B) - (A + B) = 3A + B. The common difference dd is u2−u1u_2 - u_1. d=(3A+B)−(A+B)=2Ad = (3A + B) - (A + B) = 2A. Alternatively, one can find the general term un=Sn−Sn−1=(An2+Bn)−(A(n−1)2+B(n−1))=(An2+Bn)−(An2−2An+A+Bn−B)=2An−A+Bu_n = S_n - S_{n-1} = (An^2+Bn) - (A(n-1)^2+B(n-1)) = (An^2+Bn) - (An^2-2An+A+Bn-B) = 2An - A + B. This is a linear expression in nn, and the common difference is the coefficient of nn, which is 2A2A. Distractor Explanations: A. A: A possible guess based on the quadratic term. B. B: A possible guess based on the linear term. C. A + B: This expression represents the first term of the sequence, u1u_1.

Question 12

In an arithmetic series, the sum of the first 6 terms is 78, and the sum of the first 10 terms is 230. Find the third term, u3u_3.

  1. 0.5
  2. 5
  3. 10.5 (correct answer)
  4. 15.5
Explanation: We can set up a system of two linear equations with variables u1u_1 and dd. Equation 1 from S6=78S_6=78: S6=62(2u1+(6−1)d)=3(2u1+5d)=78⇒2u1+5d=26S_6 = \frac{6}{2}(2u_1 + (6-1)d) = 3(2u_1 + 5d) = 78 \Rightarrow 2u_1 + 5d = 26. Equation 2 from S10=230S_{10}=230: S10=102(2u1+(10−1)d)=5(2u1+9d)=230⇒2u1+9d=46S_{10} = \frac{10}{2}(2u_1 + (10-1)d) = 5(2u_1 + 9d) = 230 \Rightarrow 2u_1 + 9d = 46. Subtract Equation 1 from Equation 2: (2u1+9d)−(2u1+5d)=46−26(2u_1 + 9d) - (2u_1 + 5d) = 46 - 26 4d=20⇒d=54d = 20 \Rightarrow d=5. Substitute d=5d=5 into Equation 1: 2u1+5(5)=26⇒2u1+25=26⇒2u1=1⇒u1=0.52u_1 + 5(5) = 26 \Rightarrow 2u_1 + 25 = 26 \Rightarrow 2u_1 = 1 \Rightarrow u_1 = 0.5. Finally, find the third term: u3=u1+2d=0.5+2(5)=0.5+10=10.5u_3 = u_1 + 2d = 0.5 + 2(5) = 0.5 + 10 = 10.5. Distractor Explanations: A. 0.5: This is the value of the first term u1u_1, not the third term. B. 5: This is the value of the common difference dd, not the third term. D. 15.5: This results from a sign error when solving for dd. If one found d=−5d=-5, then 2u1−25=26⇒2u1=51⇒u1=25.52u_1-25=26 \Rightarrow 2u_1=51 \Rightarrow u_1=25.5. Then u3=25.5+2(−5)=15.5u_3 = 25.5 + 2(-5) = 15.5.

Question 13

An arithmetic sequence has first term u1=5u_1=5 and common difference d=3d=3. Find the sum of the terms from the 11th to the 20th, inclusive.

  1. 450
  2. 485 (correct answer)
  3. 517
  4. 670
Explanation: The sum required is u11+u12+⋯+u20u_{11} + u_{12} + \dots + u_{20}. This can be calculated as the sum of the first 20 terms minus the sum of the first 10 terms (S20−S10S_{20} - S_{10}). First, calculate S20S_{20}: S20=202(2(5)+(20−1)(3))=10(10+19×3)=10(10+57)=670S_{20} = \frac{20}{2}(2(5) + (20-1)(3)) = 10(10 + 19 \times 3) = 10(10 + 57) = 670. Next, calculate S10S_{10}: S10=102(2(5)+(10−1)(3))=5(10+9×3)=5(10+27)=5(37)=185S_{10} = \frac{10}{2}(2(5) + (10-1)(3)) = 5(10 + 9 \times 3) = 5(10 + 27) = 5(37) = 185. The required sum is S20−S10=670−185=485S_{20} - S_{10} = 670 - 185 = 485. Alternatively, we can treat this as a new arithmetic series with n=10n=10 terms. The first term is a1=u11=5+(11−1)(3)=35a_1 = u_{11} = 5 + (11-1)(3) = 35. The last term is a10=u20=5+(20−1)(3)=62a_{10} = u_{20} = 5 + (20-1)(3) = 62. The sum is 102(35+62)=5(97)=485\frac{10}{2}(35 + 62) = 5(97) = 485. Distractor Explanations: A. 450: This is the result of an off-by-one error in the range, calculating S20−S11S_{20} - S_{11}. C. 517: This is the result of another common off-by-one error, calculating S20−S9S_{20} - S_9. D. 670: This is the sum of the first 20 terms, S20S_{20}, without subtracting the first 10.

Question 14

In an arithmetic sequence, the sum of the second and sixth terms is 32. The product of the first and fourth terms is 64. Find the common difference, dd.

  1. 2
  2. 4 (correct answer)
  3. 8
  4. 16
Explanation: Let the first term be u1u_1 and the common difference be dd. From the given information, we can write two equations:
  1. u2+u6=32u_2 + u_6 = 32 (u1+d)+(u1+5d)=32(u_1 + d) + (u_1 + 5d) = 32 2u1+6d=322u_1 + 6d = 32 Dividing by 2, we get u1+3d=16u_1 + 3d = 16. Note that the left side is the formula for the fourth term, u4u_4. So, u4=16u_4 = 16.
  2. u1⋅u4=64u_1 \cdot u_4 = 64 Substitute u4=16u_4 = 16 into the second equation: u1⋅16=64u_1 \cdot 16 = 64 u1=4u_1 = 4. Now use the equation u1+3d=16u_1 + 3d = 16 to find dd: 4+3d=164 + 3d = 16 3d=123d = 12 d=4d = 4.
Distractor Explanations: A. 2: This value for u1u_1 is obtained if one incorrectly simplifies 2u1+6d=322u_1+6d=32 to u1+3d=32u_1+3d=32, then substitutes into u1(u1+3d)=64u_1(u_1+3d)=64 to get u1(32)=64⇒u1=2u_1(32)=64 \Rightarrow u_1=2. C. 8: This is the value of the second term, u2=u1+d=4+4=8u_2 = u_1+d = 4+4=8. D. 16: This is the value of the fourth term, u4u_4.

Question 15

An arithmetic series with first term u1=7u_1 = 7 and common difference d=2d=2 has a sum of 40. Find the number of terms, nn.

  1. 4 (correct answer)
  2. 5
  3. 8
  4. 10
Explanation: We use the formula for the sum of an arithmetic series, Sn=n2(2u1+(n−1)d)S_n = \frac{n}{2}(2u_1 + (n-1)d), and solve for nn. Given Sn=40S_n=40, u1=7u_1=7, and d=2d=2: 40=n2(2(7)+(n−1)2)40 = \frac{n}{2}(2(7) + (n-1)2) 40=n2(14+2n−2)40 = \frac{n}{2}(14 + 2n - 2) 40=n2(2n+12)40 = \frac{n}{2}(2n + 12) 40=n(n+6)40 = n(n + 6) 40=n2+6n40 = n^2 + 6n n2+6n−40=0n^2 + 6n - 40 = 0. We can factor this quadratic equation: (n+10)(n−4)=0(n+10)(n-4) = 0. The possible solutions are n=−10n=-10 and n=4n=4. Since the number of terms must be a positive integer, we have n=4n=4. Check: the first 4 terms are 7, 9, 11, 13. Their sum is 7+9+11+13=407+9+11+13 = 40. Distractor Explanations: B. 5: A plausible guess or result of a calculation error. C. 8: A possible error in solving the quadratic equation. D. 10: This is the magnitude of the negative root of the quadratic equation, which is rejected. A student might make a sign error in factoring, such as (n−10)(n+4)=0(n-10)(n+4)=0, and choose the positive root.

Question 16

An arithmetic sequence has a first term of 83 and a common difference of -4. What is the sum of all the positive terms of this sequence?

  1. 900
  2. 902
  3. 903 (correct answer)
  4. 924
Explanation: First, we need to find how many terms of the sequence are positive. Let unu_n be the nn-th term. The formula is un=u1+(n−1)du_n = u_1 + (n-1)d. We want to find nn such that un>0u_n > 0. 83+(n−1)(−4)>083 + (n-1)(-4) > 0 83−4n+4>083 - 4n + 4 > 0 87>4n87 > 4n n<874=21.75n < \frac{87}{4} = 21.75. Since nn must be an integer, the terms u1,u2,…,u21u_1, u_2, \dots, u_{21} are positive. So there are 21 positive terms. Now, we find the sum of these 21 terms. We can use the formula Sn=n2(2u1+(n−1)d)S_n = \frac{n}{2}(2u_1 + (n-1)d). S21=212(2(83)+(21−1)(−4))=212(166+20(−4))=212(166−80)=212(86)=21×43=903S_{21} = \frac{21}{2}(2(83) + (21-1)(-4)) = \frac{21}{2}(166 + 20(-4)) = \frac{21}{2}(166 - 80) = \frac{21}{2}(86) = 21 \times 43 = 903. Distractor Explanations: A. 900: This is the sum of the first 20 terms, S20=202(166+19(−4))=10(166−76)=10(90)=900S_{20} = \frac{20}{2}(166+19(-4)) = 10(166-76) = 10(90)=900, which is an incorrect count of positive terms. B. 902: This is a very common error. A student might incorrectly determine that there are 22 positive terms (by rounding 21.75 up). The 22nd term is u22=83+21(−4)=−1u_{22} = 83 + 21(-4) = -1. The sum of the first 22 terms is S22=S21+u22=903−1=902S_{22} = S_{21} + u_{22} = 903 - 1 = 902. D. 924: This could be a simple calculation error, for example 21×44=92421 \times 44 = 924.

Question 17

The first three terms of an arithmetic sequence are 3k+13k+1, k+5k+5, and (2k). Find the sum of the first 15 terms of this sequence.

  1. -75
  2. -60 (correct answer)
  3. -18
  4. 360
Explanation: In an arithmetic sequence, the common difference between consecutive terms is constant. Therefore, u2−u1=u3−u2u_2 - u_1 = u_3 - u_2. (k+5)−(3k+1)=(2k)−(k+5)(k+5) - (3k+1) = (2k) - (k+5) −2k+4=k−5-2k + 4 = k - 5 9=3k⇒k=39 = 3k \Rightarrow k=3. Now substitute k=3k=3 to find the first term and common difference. u1=3(3)+1=10u_1 = 3(3) + 1 = 10 u2=3+5=8u_2 = 3 + 5 = 8 u3=2(3)=6u_3 = 2(3) = 6 The common difference is d=u2−u1=8−10=−2d = u_2 - u_1 = 8 - 10 = -2. Now find the sum of the first 15 terms using Sn=n2(2u1+(n−1)d)S_n = \frac{n}{2}(2u_1 + (n-1)d). S15=152(2(10)+(15−1)(−2))=152(20+14(−2))=152(20−28)=152(−8)=−60S_{15} = \frac{15}{2}(2(10) + (15-1)(-2)) = \frac{15}{2}(20 + 14(-2)) = \frac{15}{2}(20 - 28) = \frac{15}{2}(-8) = -60. Distractor Explanations: A. -75: This results from using ndnd instead of (n−1)d(n-1)d in the sum formula: S15=152(20+15(−2))=152(−10)=−75S_{15} = \frac{15}{2}(20 + 15(-2)) = \frac{15}{2}(-10) = -75. C. -18: This is the value of the 15th term u15=10+14(−2)=−18u_{15} = 10 + 14(-2) = -18, not the sum. D. 360: This results from a sign error for the common difference, using d=2d=2 instead of d=−2d=-2: S15=152(20+14(2))=152(48)=360S_{15} = \frac{15}{2}(20 + 14(2)) = \frac{15}{2}(48) = 360.

Question 18

The first term of an arithmetic series is 8 and the common difference is 3. Let SnS_n be the sum of the first nn terms. Find the minimum value of nn for which Sn>200S_n > 200.

  1. 9
  2. 10 (correct answer)
  3. 11
  4. 66
Explanation: We need to find the smallest integer nn that satisfies the inequality Sn>200S_n > 200. The formula for the sum is Sn=n2(2u1+(n−1)d)S_n = \frac{n}{2}(2u_1 + (n-1)d). Given u1=8u_1=8 and d=3d=3, we have: n2(2(8)+(n−1)3)>200\frac{n}{2}(2(8) + (n-1)3) > 200. n2(16+3n−3)>200\frac{n}{2}(16 + 3n - 3) > 200 n2(3n+13)>200\frac{n}{2}(3n + 13) > 200 3n2+13n>4003n^2 + 13n > 400 3n2+13n−400>03n^2 + 13n - 400 > 0. To find when this inequality holds, we first find the positive root of 3n2+13n−400=03n^2 + 13n - 400 = 0. Using the quadratic formula: n=−13+132−4(3)(−400)2(3)=−13+169+48006=−13+49696n = \frac{-13 + \sqrt{13^2 - 4(3)(-400)}}{2(3)} = \frac{-13 + \sqrt{169 + 4800}}{6} = \frac{-13 + \sqrt{4969}}{6}. Since 4900=70\sqrt{4900} = 70, 4969≈70.5\sqrt{4969} \approx 70.5. So, n≈−13+70.56=57.56≈9.58n \approx \frac{-13 + 70.5}{6} = \frac{57.5}{6} \approx 9.58. Since nn must be an integer and the quadratic opens upwards, we need n>9.58n > 9.58. The smallest integer nn is 10. We can check: S9=92(16+8(3))=92(40)=180S_9 = \frac{9}{2}(16+8(3)) = \frac{9}{2}(40) = 180, which is not > 200. S10=102(16+9(3))=5(16+27)=5(43)=215S_{10} = \frac{10}{2}(16+9(3)) = 5(16+27) = 5(43) = 215, which is > 200. So, n=10n=10 is correct. Distractor Explanations: A. 9: This would be the answer if one incorrectly rounds 9.589.58 down. C. 11: This is the next integer after the correct answer, possibly chosen due to a calculation error. D. 66: This is the answer to the different question of finding the first term unu_n to exceed 200: 8+(n−1)3>200⇒3(n−1)>192⇒n−1>64⇒n>658 + (n-1)3 > 200 \Rightarrow 3(n-1) > 192 \Rightarrow n-1 > 64 \Rightarrow n > 65, so n=66n=66.

Question 19

Find the value of the sum ∑r=420(5r−3)\sum_{r=4}^{20} (5r-3).

  1. 912
  2. 969 (correct answer)
  3. 990
  4. 1140
Explanation: This is the sum of an arithmetic series. We need to find the number of terms, the first term, and the last term. The number of terms is n=20−4+1=17n = 20 - 4 + 1 = 17. The first term (when r=4r=4) is u1=5(4)−3=20−3=17u_1 = 5(4) - 3 = 20 - 3 = 17. The last term (when r=20r=20) is u17=5(20)−3=100−3=97u_{17} = 5(20) - 3 = 100 - 3 = 97. Using the sum formula Sn=n2(u1+un)S_n = \frac{n}{2}(u_1 + u_n): S17=172(17+97)=172(114)=17×57=969S_{17} = \frac{17}{2}(17 + 97) = \frac{17}{2}(114) = 17 \times 57 = 969. Distractor Explanations: A. 912: This result comes from a common error in calculating the number of terms as n=20−4=16n = 20 - 4 = 16. Then S=162(17+97)=8(114)=912S = \frac{16}{2}(17+97) = 8(114) = 912. C. 990: This is the sum from r=1r=1 to r=20r=20. First term 5(1)−3=25(1)-3=2, last term 97, n=20n=20. S=202(2+97)=10(99)=990S = \frac{20}{2}(2+97) = 10(99) = 990. D. 1140: This result comes from incorrectly using n=20n=20 for the number of terms, instead of n=17n=17. S=202(17+97)=10(114)=1140S = \frac{20}{2}(17+97) = 10(114) = 1140.

Question 20

The sum of the first nn terms of an arithmetic sequence is given by the formula Sn=An2+BnS_n = An^2 + Bn. Which expression represents the common difference of the sequence?

  1. A
  2. B
  3. A + B
  4. 2A (correct answer)
Explanation: To find the common difference, we can find the first two terms and subtract them. The first term is u1=S1u_1 = S_1. u1=A(1)2+B(1)=A+Bu_1 = A(1)^2 + B(1) = A + B. The sum of the first two terms is S2=A(2)2+B(2)=4A+2BS_2 = A(2)^2 + B(2) = 4A + 2B. The second term is u2=S2−S1u_2 = S_2 - S_1. u2=(4A+2B)−(A+B)=3A+Bu_2 = (4A + 2B) - (A + B) = 3A + B. The common difference dd is u2−u1u_2 - u_1. d=(3A+B)−(A+B)=2Ad = (3A + B) - (A + B) = 2A. Alternatively, one can find the general term un=Sn−Sn−1=(An2+Bn)−(A(n−1)2+B(n−1))=(An2+Bn)−(An2−2An+A+Bn−B)=2An−A+Bu_n = S_n - S_{n-1} = (An^2+Bn) - (A(n-1)^2+B(n-1)) = (An^2+Bn) - (An^2-2An+A+Bn-B) = 2An - A + B. This is a linear expression in nn, and the common difference is the coefficient of nn, which is 2A2A. Distractor Explanations: A. A: A possible guess based on the quadratic term. B. B: A possible guess based on the linear term. C. A + B: This expression represents the first term of the sequence, u1u_1.