Arithmetic Sequences & Series - IB Mathematics Analysis and Approaches

1. In an arithmetic sequence, the fifth term is 18 and the ninth term is 34. Find the sum of the first 20 terms.
   Answer: 800 Let the first term be $u_1$ and the common difference be $d$. We are given $u_5 = u_1 + 4d = 18$ and $u_9 = u_1 + 8d = 34$. Subtracting the first equation from the second gives: $(u_1 + 8d) - (u_1 + 4d) = 34 - 18$, which simplifies to $4d = 16$, so $d=4$. Substitute $d=4$ into the first equation: $u_1 + 4(4) = 18 \Rightarrow u_1 + 16 = 18 \Rightarrow u_1 = 2$. Now we find the sum of the first 20 terms using the formula $S_n = \frac{n}{2}(2u_1 + (n-1)d)$. $S_{20} = \frac{20}{2}(2(2) + (20-1)(4)) = 10(4 + 19 \times 4) = 10(4 + 76) = 10(80) = 800$. Distractor Explanations: A. 720: This results from a sign error when finding $u_1$. If one calculates $u_1 = 18 - 4(4) = 2$ but then makes an error such as $u_1 = -2$, the sum would be $S_{20} = 10(2(-2) + 19(4)) = 10(-4+76)=720$. C. 840: This is a common off-by-one error, using $nd$ instead of $(n-1)d$ in the sum formula: $S_{20} = 10(2(2) + 20(4)) = 10(4+80) = 840$. D. 1120: This results from mistakenly using $u_5=18$ as the first term $u_1$: $S_{20} = 10(2(18) + 19(4)) = 10(36+76) = 1120$.
2. 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?
   Answer: 903 First, we need to find how many terms of the sequence are positive. Let $u_n$ be the $n$-th term. The formula is $u_n = u_1 + (n-1)d$. We want to find $n$ such that $u_n > 0$. $83 + (n-1)(-4) > 0$ $83 - 4n + 4 > 0$ $87 > 4n$ $n < \frac{87}{4} = 21.75$. Since $n$ must be an integer, the terms $u_1, u_2, \dots, u_{21}$ are positive. So there are 21 positive terms. Now, we find the sum of these 21 terms. We can use the formula $S_n = \frac{n}{2}(2u_1 + (n-1)d)$. $S_{21} = \frac{21}{2}(2(83) + (21-1)(-4)) = \frac{21}{2}(166 + 20(-4)) = \frac{21}{2}(166 - 80) = \frac{21}{2}(86) = 21 \times 43 = 903$. Distractor Explanations: A. 900: This is the sum of the first 20 terms, $S_{20} = \frac{20}{2}(166+19(-4)) = 10(166-76) = 10(90)=900$, 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 $u_{22} = 83 + 21(-4) = -1$. The sum of the first 22 terms is $S_{22} = S_{21} + u_{22} = 903 - 1 = 902$. D. 924: This could be a simple calculation error, for example $21 \times 44 = 924$.
3. Find the total revenue from a full house.
   Answer: $46,900 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 $u_1=20$ and $d=2$. Part 1: Seats and revenue for the first 10 rows ($n=10$). Number of seats: $S_{10} = \frac{10}{2}(2(20) + (10-1)2) = 5(40 + 18) = 5(58) = 290$ seats. Revenue: $290 \times \$50 = \$14,500$. Part 2: Seats and revenue for the remaining rows (rows 11 to 25, which is $25-11+1=15$ rows). Total seats in all 25 rows: $S_{25} = \frac{25}{2}(2(20) + (25-1)2) = \frac{25}{2}(40 + 48) = \frac{25}{2}(88) = 25 \times 44 = 1100$ seats. Seats in rows 11-25: $S_{25} - S_{10} = 1100 - 290 = 810$ seats. Revenue: $810 \times \$40 = \$32,400$. Part 3: Total revenue. Total Revenue = $\$14,500 + \$32,400 = \$46,900$. Distractor Explanations: A. $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: $290 \times \$40 + 810 \times \$50 = \$11,600 + \$40,500 = \$52,100$.
4. The sum of three consecutive terms in an arithmetic sequence is 36. The product of these three terms is 1428. Find the smallest of these three terms.
   Answer: 7 Let the three consecutive terms be $a-d$, $a$, and $a+d$, where $a$ is the middle term and $d$ is the common difference. The sum is $(a-d) + a + (a+d) = 3a$. We are given the sum is 36, so $3a = 36 \Rightarrow a = 12$. The terms are $12-d$, 12, and $12+d$. The product is $(12-d)(12)(12+d) = 1428$. Divide by 12: $(12-d)(12+d) = \frac{1428}{12} = 119$. Using the difference of squares formula, $12^2 - d^2 = 119$. $144 - d^2 = 119$. $d^2 = 144 - 119 = 25$. So, $d = \pm 5$. If $d=5$, the terms are $12-5, 12, 12+5$, which are 7, 12, 17. If $d=-5$, the terms are $12-(-5), 12, 12-5$, which are 17, 12, 7. In both cases, the set of terms is {7, 12, 17}. The smallest of these is 7. Distractor Explanations: A. -13: This results from an algebraic error in the product expansion. If a student writes $12^2-d=119$ instead of $12^2-d^2=119$, they get $144-d=119 \Rightarrow d=25$. The terms would be $12-25, 12, 12+25$, which are $-13, 12, 37$, making the smallest -13. C. 12: This is the middle term of the sequence. D. 17: This is the largest term of the sequence.
5. Find the value of $n$ for which the $n$-th terms of the two sequences are equal.
   Answer: 4 Let $A_n$ be the $n$-th term of sequence A, and $B_n$ be the $n$-th term of sequence B. The formula for the $n$-th term is $u_n = u_1 + (n-1)d$. For sequence A: $A_n = 5 + (n-1)2 = 5 + 2n - 2 = 2n + 3$. For sequence B: $B_n = 2 + (n-1)3 = 2 + 3n - 3 = 3n - 1$. We want to find $n$ such that $A_n = B_n$. $2n + 3 = 3n - 1$ Subtract $2n$ from both sides: $3 = n - 1$. Add 1 to both sides: $4 = n$. So, the 4th terms are equal. Distractor Explanations: A. 3: This results from using $nd$ instead of $(n-1)d$: $5+2n = 2+3n \Rightarrow n=3$. C. 5: A possible calculation error. D. 7: This is the value of $n$ for which the sums of the two sequences are equal, $S_{A,n} = S_{B,n}$. $S_{A,n} = \frac{n}{2}(10+(n-1)2) = n(n+4)$. $S_{B,n} = \frac{n}{2}(4+(n-1)3) = \frac{n}{2}(3n+1)$. Setting them equal: $n^2+4n = \frac{3n^2+n}{2} \Rightarrow 2n^2+8n = 3n^2+n \Rightarrow n^2-7n=0 \Rightarrow n(n-7)=0$. Since $n>0$, $n=7$.
6. Find the value of the sum $\sum_{r=4}^{20} (5r-3)$.
   Answer: 969 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 $n = 20 - 4 + 1 = 17$. The first term (when $r=4$) is $u_1 = 5(4) - 3 = 20 - 3 = 17$. The last term (when $r=20$) is $u_{17} = 5(20) - 3 = 100 - 3 = 97$. Using the sum formula $S_n = \frac{n}{2}(u_1 + u_n)$: $S_{17} = \frac{17}{2}(17 + 97) = \frac{17}{2}(114) = 17 \times 57 = 969$. Distractor Explanations: A. 912: This result comes from a common error in calculating the number of terms as $n = 20 - 4 = 16$. Then $S = \frac{16}{2}(17+97) = 8(114) = 912$. C. 990: This is the sum from $r=1$ to $r=20$. First term $5(1)-3=2$, last term 97, $n=20$. $S = \frac{20}{2}(2+97) = 10(99) = 990$. D. 1140: This result comes from incorrectly using $n=20$ for the number of terms, instead of $n=17$. $S = \frac{20}{2}(17+97) = 10(114) = 1140$.
7. 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, $u_6$?
   Answer: 13 First, find the common difference, $d$, using the sum formula $S_n = \frac{n}{2}(2u_1 + (n-1)d)$. We have $S_{10}=120$, $u_1=3$, and $n=10$. $120 = \frac{10}{2}(2(3) + (10-1)d)$ $120 = 5(6 + 9d)$ $24 = 6 + 9d$ $18 = 9d \Rightarrow d=2$. Now, find the sixth term using the formula $u_n = u_1 + (n-1)d$. $u_6 = 3 + (6-1)(2) = 3 + 5(2) = 3 + 10 = 13$. Distractor Explanations: A. -7: This is the result of a sign error when finding $d$. If $9d = -18 \Rightarrow d=-2$, then $u_6 = 3 + 5(-2) = -7$. B. 2: This is the value of the common difference $d$, not the sixth term. D. 15: This is a common off-by-one error, using $nd$ instead of $(n-1)d$ to find the term: $u_6 = u_1 + 6d = 3 + 6(2) = 15$.
8. The first three terms of an arithmetic sequence are $3k+1$, $k+5$, and (2k). Find the sum of the first 15 terms of this sequence.
   Answer: -60 In an arithmetic sequence, the common difference between consecutive terms is constant. Therefore, $u_2 - u_1 = u_3 - u_2$. $(k+5) - (3k+1) = (2k) - (k+5)$ $-2k + 4 = k - 5$ $9 = 3k \Rightarrow k=3$. Now substitute $k=3$ to find the first term and common difference. $u_1 = 3(3) + 1 = 10$ $u_2 = 3 + 5 = 8$ $u_3 = 2(3) = 6$ The common difference is $d = u_2 - u_1 = 8 - 10 = -2$. Now find the sum of the first 15 terms using $S_n = \frac{n}{2}(2u_1 + (n-1)d)$. $S_{15} = \frac{15}{2}(2(10) + (15-1)(-2)) = \frac{15}{2}(20 + 14(-2)) = \frac{15}{2}(20 - 28) = \frac{15}{2}(-8) = -60$. Distractor Explanations: A. -75: This results from using $nd$ instead of $(n-1)d$ in the sum formula: $S_{15} = \frac{15}{2}(20 + 15(-2)) = \frac{15}{2}(-10) = -75$. C. -18: This is the value of the 15th term $u_{15} = 10 + 14(-2) = -18$, not the sum. D. 360: This results from a sign error for the common difference, using $d=2$ instead of $d=-2$: $S_{15} = \frac{15}{2}(20 + 14(2)) = \frac{15}{2}(48) = 360$.
9. In an arithmetic progression, the sum of the first $p$ terms is equal to the sum of the first $q$ terms, where $p \neq q$. Find the sum of the first $p+q$ terms.
   Answer: 0 When you encounter arithmetic progressions where two different partial sums are equal, you're dealing with a special symmetry property that reveals important information about the sequence structure. Let's work with the general formula for the sum of the first $n$ terms of an arithmetic progression: $S_n = \frac{n}{2}[2a + (n-1)d]$, where $a$ is the first term and $d$ is the common difference. Given that $S_p = S_q$ where $p \neq q$, we have: $\frac{p}{2}[2a + (p-1)d] = \frac{q}{2}[2a + (q-1)d]$ Expanding and simplifying: $p[2a + (p-1)d] = q[2a + (q-1)d]$ $2ap + p(p-1)d = 2aq + q(q-1)d$ $2a(p-q) + d[p(p-1) - q(q-1)] = 0$ $2a(p-q) + d(p^2-q^2-p+q) = 0$ $2a(p-q) + d(p-q)(p+q-1) = 0$ $(p-q)[2a + d(p+q-1)] = 0$ Since $p \neq q$, we have $p-q \neq 0$, so: $2a + d(p+q-1) = 0$ Now for $S_{p+q}$: $S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d] = \frac{p+q}{2} \cdot 0 = 0$ Choice A) $-(p+q)$ incorrectly assumes a negative relationship. Choice B) $p+q$ might tempt you to think the sum equals the number of terms. Choice C) $1$ has no mathematical basis here. Choice D) $0$ is correct. Key insight: When two different partial sums of an arithmetic progression are equal, the terms are arranged symmetrically around zero, making $S_{p+q} = 0$ inevitable. Remember this elegant symmetry property—it's a powerful shortcut on IB exams.
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, $d$.
    Answer: 4 Let the first term be $u_1$ and the common difference be $d$. From the given information, we can write two equations: 1) $u_2 + u_6 = 32$ $(u_1 + d) + (u_1 + 5d) = 32$ $2u_1 + 6d = 32$ Dividing by 2, we get $u_1 + 3d = 16$. Note that the left side is the formula for the fourth term, $u_4$. So, $u_4 = 16$. 2) $u_1 \cdot u_4 = 64$ Substitute $u_4 = 16$ into the second equation: $u_1 \cdot 16 = 64$ $u_1 = 4$. Now use the equation $u_1 + 3d = 16$ to find $d$: $4 + 3d = 16$ $3d = 12$ $d = 4$. Distractor Explanations: A. 2: This value for $u_1$ is obtained if one incorrectly simplifies $2u_1+6d=32$ to $u_1+3d=32$, then substitutes into $u_1(u_1+3d)=64$ to get $u_1(32)=64 \Rightarrow u_1=2$. C. 8: This is the value of the second term, $u_2 = u_1+d = 4+4=8$. D. 16: This is the value of the fourth term, $u_4$.
11. The sum of the first $n$ terms of a sequence is given by $S_n = 2n^2 - 5n$. Find the 10th term, $u_{10}$.
    Answer: 33 The $n$-th term of a sequence can be found using the formula $u_n = S_n - S_{n-1}$ for $n \ge 2$. To find the 10th term, we calculate $u_{10} = S_{10} - S_9$. First, calculate $S_{10}$: $S_{10} = 2(10)^2 - 5(10) = 2(100) - 50 = 200 - 50 = 150$. Next, calculate $S_9$: $S_9 = 2(9)^2 - 5(9) = 2(81) - 45 = 162 - 45 = 117$. Now, find $u_{10}$: $u_{10} = S_{10} - S_9 = 150 - 117 = 33$. Distractor Explanations: A. 4: This is the common difference of the sequence. $u_1 = S_1 = -3$, $u_2 = S_2 - S_1 = (2(4)-5(2)) - (-3) = -2 - (-3) = 1$. So $d = u_2 - u_1 = 1 - (-3) = 4$. C. 117: This is the value of $S_9$, the sum of the first nine terms. D. 150: This is the value of $S_{10}$, the sum of the first ten terms. This is a common error made by students who confuse the sum with the term itself.
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, $u_3$.
    Answer: 10.5 We can set up a system of two linear equations with variables $u_1$ and $d$. Equation 1 from $S_6=78$: $S_6 = \frac{6}{2}(2u_1 + (6-1)d) = 3(2u_1 + 5d) = 78 \Rightarrow 2u_1 + 5d = 26$. Equation 2 from $S_{10}=230$: $S_{10} = \frac{10}{2}(2u_1 + (10-1)d) = 5(2u_1 + 9d) = 230 \Rightarrow 2u_1 + 9d = 46$. Subtract Equation 1 from Equation 2: $(2u_1 + 9d) - (2u_1 + 5d) = 46 - 26$ $4d = 20 \Rightarrow d=5$. Substitute $d=5$ into Equation 1: $2u_1 + 5(5) = 26 \Rightarrow 2u_1 + 25 = 26 \Rightarrow 2u_1 = 1 \Rightarrow u_1 = 0.5$. Finally, find the third term: $u_3 = u_1 + 2d = 0.5 + 2(5) = 0.5 + 10 = 10.5$. Distractor Explanations: A. 0.5: This is the value of the first term $u_1$, not the third term. B. 5: This is the value of the common difference $d$, not the third term. D. 15.5: This results from a sign error when solving for $d$. If one found $d=-5$, then $2u_1-25=26 \Rightarrow 2u_1=51 \Rightarrow u_1=25.5$. Then $u_3 = 25.5 + 2(-5) = 15.5$.
13. In which year will the company's total profit since it started first exceed $2,000,000?
    Answer: 13th 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: $S_n = \frac{n}{2}[2a + (n-1)d]$ where $a = 120,000$ and $d = 8,000$. 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 $S_n > 2,000,000$, so: $4,000n(58 + n) > 2,000,000$ $n(58 + n) > 500$ $n^2 + 58n - 500 > 0$ Using the quadratic formula: $n = \frac{-58 + \sqrt{58^2 + 4(500)}}{2} \approx 7.6$ Since we need the first year to exceed $2,000,000, we test: $S_{12} = 4,000(12)(70) = 3,360,000$ and $S_{13} = 4,000(13)(71) = 3,692,000$. 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.
14. An arithmetic series with first term $u_1 = 7$ and common difference $d=2$ has a sum of 40. Find the number of terms, $n$.
    Answer: 4 We use the formula for the sum of an arithmetic series, $S_n = \frac{n}{2}(2u_1 + (n-1)d)$, and solve for $n$. Given $S_n=40$, $u_1=7$, and $d=2$: $40 = \frac{n}{2}(2(7) + (n-1)2)$ $40 = \frac{n}{2}(14 + 2n - 2)$ $40 = \frac{n}{2}(2n + 12)$ $40 = n(n + 6)$ $40 = n^2 + 6n$ $n^2 + 6n - 40 = 0$. We can factor this quadratic equation: $(n+10)(n-4) = 0$. The possible solutions are $n=-10$ and $n=4$. Since the number of terms must be a positive integer, we have $n=4$. Check: the first 4 terms are 7, 9, 11, 13. Their sum is $7+9+11+13 = 40$. 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 $(n-10)(n+4)=0$, and choose the positive root.
15. Let $u_n$ be an arithmetic sequence with common difference $d$. Find the value of the sum $S = u_1 - u_2 + u_3 - u_4 + \dots + u_{29} - u_{30}$.
    Answer: -15d When you encounter alternating sums with arithmetic sequences, look for patterns by grouping consecutive terms. This approach often reveals surprising simplifications. In an arithmetic sequence, each term can be written as $u_n = u_1 + (n-1)d$. Let's group the sum into pairs: $(u_1 - u_2) + (u_3 - u_4) + \dots + (u_{29} - u_{30})$. For any pair $(u_{2k-1} - u_{2k})$, we get: $u_{2k-1} - u_{2k} = [u_1 + (2k-2)d] - [u_1 + (2k-1)d] = -d$ Since we have 30 terms total, there are 15 such pairs, and each pair contributes $-d$ to the sum. Therefore: $S = 15 \times (-d) = -15d$. Answer choice A (-30d) likely comes from incorrectly thinking there are 30 pairs instead of recognizing that 30 terms form 15 pairs. Answer choice B ($u_1 - 15d$) might result from trying to use the arithmetic series formula inappropriately or from grouping terms incorrectly around $u_1$. Answer choice C (15d) gets the magnitude right but misses the crucial negative sign that comes from the alternating pattern. The correct answer is D: $-15d$. Study tip: For alternating sums in arithmetic sequences, always group consecutive terms in pairs first. The pattern within each pair often simplifies dramatically, making the entire sum much easier to calculate than working with individual terms.
16. The sum of the first $n$ terms of an arithmetic sequence is given by the formula $S_n = An^2 + Bn$. Which expression represents the common difference of the sequence?
    Answer: 2A To find the common difference, we can find the first two terms and subtract them. The first term is $u_1 = S_1$. $u_1 = A(1)^2 + B(1) = A + B$. The sum of the first two terms is $S_2 = A(2)^2 + B(2) = 4A + 2B$. The second term is $u_2 = S_2 - S_1$. $u_2 = (4A + 2B) - (A + B) = 3A + B$. The common difference $d$ is $u_2 - u_1$. $d = (3A + B) - (A + B) = 2A$. Alternatively, one can find the general term $u_n = S_n - S_{n-1} = (An^2+Bn) - (A(n-1)^2+B(n-1)) = (An^2+Bn) - (An^2-2An+A+Bn-B) = 2An - A + B$. This is a linear expression in $n$, and the common difference is the coefficient of $n$, which is $2A$. 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, $u_1$.
17. The first term of an arithmetic series is 8 and the common difference is 3. Let $S_n$ be the sum of the first $n$ terms. Find the minimum value of $n$ for which $S_n > 200$.
    Answer: 10 We need to find the smallest integer $n$ that satisfies the inequality $S_n > 200$. The formula for the sum is $S_n = \frac{n}{2}(2u_1 + (n-1)d)$. Given $u_1=8$ and $d=3$, we have: $\frac{n}{2}(2(8) + (n-1)3) > 200$. $\frac{n}{2}(16 + 3n - 3) > 200$ $\frac{n}{2}(3n + 13) > 200$ $3n^2 + 13n > 400$ $3n^2 + 13n - 400 > 0$. To find when this inequality holds, we first find the positive root of $3n^2 + 13n - 400 = 0$. Using the quadratic formula: $n = \frac{-13 + \sqrt{13^2 - 4(3)(-400)}}{2(3)} = \frac{-13 + \sqrt{169 + 4800}}{6} = \frac{-13 + \sqrt{4969}}{6}$. Since $\sqrt{4900} = 70$, $\sqrt{4969} \approx 70.5$. So, $n \approx \frac{-13 + 70.5}{6} = \frac{57.5}{6} \approx 9.58$. Since $n$ must be an integer and the quadratic opens upwards, we need $n > 9.58$. The smallest integer $n$ is 10. We can check: $S_9 = \frac{9}{2}(16+8(3)) = \frac{9}{2}(40) = 180$, which is not > 200. $S_{10} = \frac{10}{2}(16+9(3)) = 5(16+27) = 5(43) = 215$, which is > 200. So, $n=10$ is correct. Distractor Explanations: A. 9: This would be the answer if one incorrectly rounds $9.58$ 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 $u_n$ to exceed 200: $8 + (n-1)3 > 200 \Rightarrow 3(n-1) > 192 \Rightarrow n-1 > 64 \Rightarrow n > 65$, so $n=66$.
18. An arithmetic sequence has first term $u_1=5$ and common difference $d=3$. Find the sum of the terms from the 11th to the 20th, inclusive.
    Answer: 485 The sum required is $u_{11} + u_{12} + \dots + u_{20}$. This can be calculated as the sum of the first 20 terms minus the sum of the first 10 terms ($S_{20} - S_{10}$). First, calculate $S_{20}$: $S_{20} = \frac{20}{2}(2(5) + (20-1)(3)) = 10(10 + 19 \times 3) = 10(10 + 57) = 670$. Next, calculate $S_{10}$: $S_{10} = \frac{10}{2}(2(5) + (10-1)(3)) = 5(10 + 9 \times 3) = 5(10 + 27) = 5(37) = 185$. The required sum is $S_{20} - S_{10} = 670 - 185 = 485$. Alternatively, we can treat this as a new arithmetic series with $n=10$ terms. The first term is $a_1 = u_{11} = 5 + (11-1)(3) = 35$. The last term is $a_{10} = u_{20} = 5 + (20-1)(3) = 62$. The sum is $\frac{10}{2}(35 + 62) = 5(97) = 485$. Distractor Explanations: A. 450: This is the result of an off-by-one error in the range, calculating $S_{20} - S_{11}$. C. 517: This is the result of another common off-by-one error, calculating $S_{20} - S_9$. D. 670: This is the sum of the first 20 terms, $S_{20}$, without subtracting the first 10.
19. The sum of the first $n$ consecutive positive odd integers is 1225. Find the value of $n$.
    Answer: 35 The sequence of consecutive positive odd integers is 1, 3, 5, 7, ... This is an arithmetic sequence with first term $u_1 = 1$ and common difference $d = 2$. The sum of the first $n$ terms is given by $S_n = \frac{n}{2}(2u_1 + (n-1)d)$. Substituting the values, we get $S_n = \frac{n}{2}(2(1) + (n-1)2) = \frac{n}{2}(2 + 2n - 2) = \frac{n}{2}(2n) = n^2$. We are given that the sum is 1225, so we need to solve $n^2 = 1225$. Taking the square root, $n = \sqrt{1225}$. Since the number ends in 5, its square root must end in 5. $30^2=900$ and $40^2=1600$, so we can test 35. $35^2 = 1225$. Therefore, $n=35$. Distractor Explanations: B. 49: This is the value of $n$ if the student incorrectly used the formula for the sum of the first $n$ positive integers: $\frac{n(n+1)}{2} = 1225 \Rightarrow n(n+1) = 2450$, which gives $n=49$. C. 69: This is the value of the last term in the series, $u_{35} = 1 + (35-1)2 = 1+68 = 69$, not the number of terms. D. 613: This results from confusing the formula for the $n$-th term with the sum and solving for $n$: $u_n = 2n-1 = 1225 \Rightarrow 2n=1226 \Rightarrow n=613$.
20. Four arithmetic means are inserted between -3 and 17. Find the sum of these four means.
    Answer: 28 When four arithmetic means are inserted between -3 and 17, we get a new arithmetic sequence with a total of $4+2=6$ terms. Let this sequence be $u_n$. The first term is $u_1 = -3$ and the sixth term is $u_6 = 17$. We can find the common difference $d$ using the formula $u_n = u_1 + (n-1)d$. $u_6 = u_1 + (6-1)d \Rightarrow 17 = -3 + 5d$. $20 = 5d \Rightarrow d=4$. The sequence is -3, 1, 5, 9, 13, 17. The four arithmetic means are 1, 5, 9, and 13. Their sum is $1+5+9+13 = 28$. 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 $-3+17=14$. The sum of all 6 terms is $S_6 = \frac{6}{2}(-3+17) = 3(14)=42$. The sum of the means is this total sum minus the endpoints: $42 - (-3) - 17 = 28$. Distractor Explanations: A. 4: This is the value of the common difference $d$, not the sum of the means. B. 7: This is the average of the first and last term, $(-3+17)/2 = 7$. 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.

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