Determine the general term for each arithmetic sequence.

Step 1: Determine the general term for each arithmetic sequence. The formula for the n-th term of an arithmetic sequence is a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference. For the first sequence: 15, 19, 23, 27, , 415 The first term a_1 = 15. The common difference d_1 = 19 - 15 = 4. The general term is a_n = 15 + (n-1)4 = 15 + 4n - 4 = 4n + 11. To find the number of terms in this sequence, set a_n = 415: 4n + 11 = 415 4n = 404 n = 101 So, the first sequence has 101 terms. For the second sequence: 14, 19, 24, 29, , 464 The first term a_1 = 14. The common difference d_2 = 19 - 14 = 5. The general term is a_m = 14 + (m-1)5 = 14 + 5m - 5 = 5m + 9. To find the number of terms in this sequence, set a_m = 464: 5m + 9 = 464 5m = 455 m = 91 So, the second sequence has 91 terms. Step 2: Find the first common term and the common difference of the sequence of common terms. The common terms must satisfy both general forms: 4n + 11 = 5m + 9 4n = 5m - 2 We are looking for the smallest integer values of n and m that satisfy this equation. Let's test values for m: If m=1, 5(1)-2 = 3 (not divisible by 4) If m=2, 5(2)-2 = 8 (divisible by 4, so 4n=8 n=2) When n=2, the term in the first sequence is 4(2) + 11 = 19. When m=2, the term in the second sequence is 5(2) + 9 = 19. So, the first common term is 19. The common difference of the sequence of common terms is the Least Common Multiple (LCM) of the common differences of the two original sequences. d_1 = 4 and d_2 = 5. LCM(4, 5) = 20. So, the sequence of common terms is an arithmetic sequence with first term A_1 = 19 and common difference D = 20. The general term for the common sequence is A_k = 19 + (k-1)20. Step 3: Determine the upper limit for the common terms. A common term must be present in both sequences. Therefore, it cannot exceed the last term of either sequence. The last term of the first sequence is 415. The last term of the second sequence is 464. The common terms must be less than or equal to the minimum of these two last terms: (415, 464) = 415. Step 4: Find the number of common terms. We need to find the number of terms k such that A_k 415: 19 + (k-1)20 415 (k-1)20 415 - 19 (k-1)20 396 k-1 (396)/(20) k-1 19.8 k 19.8 + 1 k 20.8 Since k must be an integer, the largest possible value for k is 20. Therefore, there are 20 common terms in the two sequences. The final answer is B) 20. That's 2 down. 3 left today — send the next one.