Missing Pattern Terms - SHSAT Math

1. A sequence starts with 41 and decreases by 6 each time. Which term in this sequence will first be negative?
   Answer: The 8th term When you encounter a sequence problem asking when something "first" happens, you're looking for the exact transition point. This arithmetic sequence starts at 41 and decreases by 6 each term, so you need to find when it crosses from positive to negative. The sequence follows the pattern: $a_n = 41 - 6(n-1)$, where $n$ is the term number. To find when the sequence first becomes negative, set up the inequality: $41 - 6(n-1) < 0$. Solving this: $41 < 6(n-1)$, so $6.83 < n-1$, which means $n > 7.83$. Since $n$ must be a whole number, the first negative term is the 8th term. You can verify by calculating: the 7th term is $41 - 6(6) = 5$ (still positive), and the 8th term is $41 - 6(7) = -1$ (first negative term). Choice A (9th term) would give you $41 - 6(8) = -7$, which is negative but not the first negative term. Choice B (7th term) equals 5, which is still positive. Choice D (6th term) equals $41 - 6(5) = 11$, also positive. These wrong answers likely come from calculation errors or misunderstanding what "first negative" means. Strategy tip: For "first time" sequence problems, always check the term before your calculated answer to confirm it's still positive (or whatever the previous condition was). This ensures you've found the exact transition point, not just any term that meets the new condition.
2. An arithmetic sequence has 15 terms. The first term is 8 and the last term is 50. What is the 8th term?
   Answer: 29 When you encounter an arithmetic sequence problem, you're working with a pattern where each term increases by the same constant difference. The key is finding that common difference and using it to locate any term in the sequence. Start with the arithmetic sequence formula: $a_n = a_1 + (n-1)d$, where $a_n$ is the nth term, $a_1$ is the first term, and $d$ is the common difference. Since you know the first term (8), last term (50), and total number of terms (15), you can find the common difference first. For the 15th term: $50 = 8 + (15-1)d$, which gives you $50 = 8 + 14d$. Solving: $42 = 14d$, so $d = 3$. Now find the 8th term: $a_8 = 8 + (8-1)(3) = 8 + 21 = 29$. The answer is C. Looking at the wrong answers: A) 31 would result from incorrectly calculating $8 + 8(3) = 32$ or making a similar error with the position. B) 27 likely comes from using $d = 2.8$ (rounding $42/15$ instead of $42/14$) or miscounting terms. D) 25 suggests using $d = 2.4$ or another computational error in finding the common difference. Remember this pattern: arithmetic sequence problems always boil down to finding the common difference first. Once you have that, you can find any term. Double-check your work by verifying that your common difference actually produces the given last term.
3. In an arithmetic sequence, the 3rd term is 17 and the 7th term is 33. What is the 12th term?
   Answer: 53 First find the common difference: between the 3rd and 7th terms there are 4 steps, so d = (33-17)/4 = 4. The first term is a₁ = 17 - 2(4) = 9. The 12th term is a₁₂ = 9 + 11(4) = 53. Choice B uses d = 3, Choice C miscalculates the first term, Choice D adds one extra common difference.
4. The first four terms of an arithmetic sequence are 7, 12, 17, 22. If this pattern continues, what is the average of the 10th and 15th terms?
   Answer: 64.5 The sequence 7, 12, 17, 22 adds 5 each time, so the $n$th term is $7 + 5(n - 1)$. 10th term: $7 + 5(9) = 52$. 15th term: $7 + 5(14) = 77$. Average: $\frac{52 + 77}{2} = \frac{129}{2} = 64.5$. The value 67 comes from using $7 + 5(10) = 57$ as the 10th term, multiplying by $n$ instead of $n - 1$. The value 62 makes the same slip in the other direction on the 15th term, using 72. The value 64 drops the half when dividing 129 by 2.
5. A pattern follows the rule that each term after the first is found by adding 6 to the previous term. If the sum of the 4th and 8th terms is 84, what is the 6th term?
   Answer: 42 Each term is 6 more than the one before. The 6th term sits exactly halfway between the 4th and 8th terms, because the 4th is two steps (12 less) before it and the 8th is two steps (12 more) after it. So the 4th term plus the 8th term equals twice the 6th term: $2 \times (\text{6th term}) = 84$, and the 6th term is 42. Check: the 4th term is $42 - 12 = 30$ and the 8th is $42 + 12 = 54$, and $30 + 54 = 84$. The first term is $30 - 18 = 12$. The value 36 is the 5th term, one step too early. The values 39 and 33 cannot be terms of this sequence at all: starting from 12 and adding 6 each time, every term is a multiple of 6, and neither odd number is.
6. An arithmetic sequence has its first term equal to -8 and its common difference equal to 5. If one term in this sequence is 47, what is the next term after 47?
   Answer: 52 In an arithmetic sequence, the next term is always found by adding the common difference. Since 47 is a term and the common difference is 5, the next term is 47 + 5 = 52. Choice B uses d = 3, Choice C uses d = 8, Choice D uses d = 2.
7. A pattern begins $\tfrac34,\;1,\;\tfrac54,\;?$. If the pattern is arithmetic, what number belongs in place of the question mark?
   Answer: $\tfrac{3}{2}$ In an arithmetic sequence the same amount is added each time. Find the common difference: $1 - \tfrac34 = \tfrac14$, and $\tfrac54 - 1 = \tfrac14$. Add it to the last term: $\tfrac54 + \tfrac14 = \tfrac64 = \tfrac32$. The value $\tfrac74$ adds $\tfrac12$ instead of $\tfrac14$, doubling the common difference. The value $\tfrac{11}{8}$ adds only $\tfrac18$, half the difference. The value $\tfrac52$ doubles $\tfrac54$, treating the pattern as multiplying rather than adding.
8. Consider the pattern: 2, 9, 16, 23, 30, ?, 44. What number belongs in the place of the question mark?
   Answer: 37 When you encounter a sequence of numbers like this, you're dealing with an arithmetic pattern where each term increases by the same amount. Your first step should be to find the common difference by subtracting consecutive terms. Let's examine the differences: $9 - 2 = 7$, $16 - 9 = 7$, $23 - 16 = 7$, and $30 - 23 = 7$. The pattern increases by 7 each time, so this is an arithmetic sequence with a common difference of 7. To find the missing number, add 7 to the term before it: $30 + 7 = 37$. You can verify this works by checking that the next term follows the pattern: $37 + 7 = 44$, which matches the given final number. Looking at the wrong answers: Choice (A) 35 would mean adding only 5 to get from 30 to 35, breaking the established pattern of adding 7. Choice (C) 38 represents adding 8, which again doesn't match the consistent difference of 7. Choice (D) 36 involves adding 6, another deviation from the pattern. Each incorrect option represents a common mistake: either miscalculating the common difference or making an arithmetic error when adding. These distractors are designed to catch students who don't systematically check their pattern. For sequence problems on the SHSAT, always identify the pattern by calculating differences between consecutive terms first. Once you find the rule, apply it consistently and double-check by verifying that your answer leads correctly to the next known term.
9. In the arithmetic sequence 5, 11, 17, 23, ..., which term has a value of 119?
   Answer: The 20th term The common difference is 6, and the first term is 5. Using aₙ = a₁ + (n-1)d, we have 119 = 5 + (n-1)6. Solving: 114 = 6(n-1), so n-1 = 19, giving n = 20. Choice B forgets to add 1 back, Choice C adds an extra term, Choice D uses an incorrect calculation.
10. The 5th term of an arithmetic sequence is 28, and the 9th term is 44. What is the 2nd term?
    Answer: 16 When you encounter arithmetic sequence problems, remember that these sequences have a constant difference between consecutive terms. Your goal is to find this common difference first, then work backward to the desired term. Since the 5th term is 28 and the 9th term is 44, you can find the common difference by recognizing that from the 5th to 9th term, you move 4 steps in the sequence. The total change is $44 - 28 = 16$, so the common difference is $d = \frac{16}{4} = 4$. Now you can work backward from the 5th term to find the 2nd term. Since each step backward decreases the term by 4, moving from the 5th term to the 2nd term (3 steps backward) gives you: $28 - 3(4) = 28 - 12 = 16$. Looking at the wrong answers: Choice A (18) represents a common error where students might add instead of subtract when moving backward, or miscalculate the number of steps. Choice B (14) could result from using an incorrect common difference of 2 instead of 4. Choice C (12) might occur if you subtract 16 directly from 28, confusing the total change between terms with the actual term value. The correct answer is D) 16. Strategy tip: For arithmetic sequences, always find the common difference first by dividing the total change by the number of steps between given terms. Then count carefully whether you're moving forward (add) or backward (subtract) from your reference point.
11. What number should replace the question mark in the arithmetic sequence $7,\;11,\;15,\;?,\;23$?
    Answer: 19 When you encounter a sequence with evenly spaced numbers, you're dealing with an arithmetic sequence where each term increases by the same amount (called the common difference). To find the common difference, subtract any term from the next term. Looking at the given sequence $7, 11, 15, ?, 23$, you can calculate: $11 - 7 = 4$ and $15 - 11 = 4$. The common difference is 4. Since each term increases by 4, the missing term must be $15 + 4 = 19$. You can verify this works by checking that $19 + 4 = 23$, which matches the final given term. Let's examine why the other choices don't work. Choice A) 17 would create a common difference of $17 - 15 = 2$ from the third to fourth term, but then $23 - 17 = 6$ from fourth to fifth term—inconsistent differences mean it's not arithmetic. Choice C) 20 gives us $20 - 15 = 5$ and $23 - 20 = 3$, again showing inconsistent differences. Choice D) 21 creates differences of $21 - 15 = 6$ and $23 - 21 = 2$, which are also inconsistent. The key strategy for arithmetic sequence problems is to always find the common difference first by subtracting consecutive terms. Once you know this constant difference, you can find any missing term by adding or subtracting appropriately. Remember that in a true arithmetic sequence, the difference between consecutive terms must be the same throughout the entire sequence.
12. The first four terms of an arithmetic sequence are $-12,\;-5,\;2,\;9$. What is the 6th term of the sequence?
    Answer: 23 When you encounter an arithmetic sequence problem, you're working with a pattern where each term increases (or decreases) by the same constant amount called the common difference. First, find the common difference by subtracting any term from the next term: $-5 - (-12) = 7$. You can verify this: $2 - (-5) = 7$ and $9 - 2 = 7$. So the common difference is 7. To find any term in an arithmetic sequence, use the formula: $a_n = a_1 + (n-1)d$, where $a_1$ is the first term, $n$ is the term number, and $d$ is the common difference. Here, $a_1 = -12$ and $d = 7$. For the 6th term: $a_6 = -12 + (6-1)(7) = -12 + 5(7) = -12 + 35 = 23$ Looking at the wrong answers: Choice (A) 16 would result from incorrectly using 4 as the multiplier instead of 5, giving $-12 + 4(7) = 16$. Choice (B) 18 might come from miscalculating the common difference as 6 instead of 7, then computing $-12 + 5(6) = 18$. Choice (D) 25 could result from using the wrong first term or making an arithmetic error in the final calculation. Strategy tip: Always verify your common difference by checking it works between multiple consecutive pairs of terms. This catches calculation errors early and ensures you're working with the correct pattern throughout the problem.

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