Use arithmetic and algebraic patterns appropriate to level - CogAT Grades 3 Plus

1. In an arithmetic sequence, the 3rd term is 17 and the 7th term is 33. What is the 12th term?
   Answer: 53 In an arithmetic sequence, $a_n$ = $a_1$ + (n-1)d where d is the common difference. From the given information: $a_3$ = 17 and $a_7$ = 33. Since $a_7$ - $a_3$ = 4d, we have 33 - 17 = 4d, so d = 4. Now find $a_1$: $a_3$ = $a_1$ + 2d = $a_1$ + 8 = 17, so $a_1$ = 9. Therefore: $a_{12}$ = $a_1$ + 11d = 9 + 11(4) = 9 + 44 = 53.
2. The sum of the first n terms of an arithmetic sequence is given by $S_n$ = $n^2$ + 2n. What is the 8th term of this sequence?
   Answer: 17 The nth term of a sequence can be found using $a_n$ = $S_n$ - $S_{n-1}$ for n ≥ 2. We have $S_8$ = $8^2$ + 2(8) = 64 + 16 = 80 and $S_7$ = $7^2$ + 2(7) = 49 + 14 = 63. Therefore, $a_8$ = $S_8$ - $S_7$ = 80 - 63 = 17. We can verify this with the general formula: $a_n$ = $S_n$ - $S_{n-1}$ = ($n^2$ + 2n) - ($(n-1)^2$ + 2(n-1)) = $n^2$ + 2n - ($n^2$ - 2n + 1 + 2n - 2) = 2n + 1. So $a_8$ = 2(8) + 1 = 17.
3. Refer to the figure below. Which number belongs in the empty oval so that the rule used in every branch is the same?
   Answer: $13$ Beginning at the top oval (4), the left branch shows 4 → 7 (add 3) → 10 (add 3). The right branch must also add 3 each time: 4 → 7, then 7 → 13. Therefore the empty oval should show $13$. Choice A (9) adds 2, choice B (11) adds 4, and choice D (15) adds 8, none of which match the consistent +3 rule.
4. In a Fibonacci-like sequence where each term is the sum of the two preceding terms, the 5th term is 21 and the 7th term is 55. What is the 9th term?
   Answer: 144 Let the terms be $a_1, a_2, a_3, ...$ where $a_n = a_{n-1} + a_{n-2}$. Given $a_5 = 21$ and $a_7 = 55$. Since $a_6 = a_7 - a_5 = 55 - 21 = 34$ (because $a_7 = a_6 + a_5$), we can find: $a_8 = a_7 + a_6 = 55 + 34 = 89$ and $a_9 = a_8 + a_7 = 89 + 55 = 144$. Choice A (132) comes from incorrectly calculating $a_6 = 32$. Choice C (156) results from the error $a_6 = 36$. Choice D (168) comes from adding incorrectly in the final step.
5. In a sequence, each term after the second is the average of all the preceding terms. If the first term is 6 and the second term is 18, what is the fourth term?
   Answer: 12 Given $a_1 = 6$ and $a_2 = 18$. The third term is the average of the first two: $a_3 = \frac{a_1 + a_2}{2} = \frac{6 + 18}{2} = 12$. The fourth term is the average of the first three: $a_4 = \frac{a_1 + a_2 + a_3}{3} = \frac{6 + 18 + 12}{3} = \frac{36}{3} = 12$. Choice A (10) comes from incorrectly calculating $a_3 = 10$ and then $a_4 = \frac{6+18+10}{3}$. Choice C (14) results from the error $a_3 = 14$. Choice D (16) comes from calculating $a_4$ as the average of just $a_2$ and $a_3$.

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