124 is the___th term of 2,5,12...

Answer:The sequence 2, 5, 12... doesn't follow a simple arithmetic or geometric pattern. Let's look at the differences between consecutive terms: - 5 - 2 = 3- 12 - 5 = 7 The differences themselves don't form a consistent pattern either. Let's try looking at the second differences: - 7 - 3 = 4 This suggests a possible quadratic pattern. A quadratic sequence has a second difference that is constant. Let's assume the general term is of the form an^2 + bn + c, where 'n' is the term number. We have three equations based on the given terms: - For n=1: a(1)² + b(1) + c = 2 => a + b + c = 2- For n=2: a(2)² + b(2) + c = 5 => 4a + 2b + c = 5- For n=3: a(3)² + b(3) + c = 12 => 9a + 3b + c = 12 Subtracting the first equation from the second and third equations, we get: - 3a + b = 3- 8a + 2b = 10 => 4a + b = 5 Subtracting the first of these new equations from the second, we get: - a = 2 Substituting a = 2 into 3a + b = 3, we get: - b = 3 - 3(2) = -3 Substituting a = 2 and b = -3 into a + b + c = 2, we get: - c = 2 - 2 + 3 = 3 Therefore, the general term is 2n^2 - 3n + 3. Let's check if this works: - n=1: 2(1)² - 3(1) + 3 = 2- n=2: 2(2)² - 3(2) + 3 = 5- n=3: 2(3)² - 3(3) + 3 = 12 Now, let's find n when the term is 124: 2n^2 - 3n + 3 = 1242n^2 - 3n - 121 = 0 Using the quadratic formula: n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} n = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-121)}}{2(2)}n = \frac{3 \pm \sqrt{9 + 968}}{4}n = \frac{3 \pm \sqrt{977}}{4}n ≈ \frac{3 \pm 31.257}{4} Since n must be a positive integer, we take the positive solution: n ≈ \frac{3 + 31.257}{4} ≈ 8.56 Since n must be an integer, there must be an error in the problem statement or the sequence provided. There is no integer solution for n. Therefore, I cannot definitively answer the question.

Answered by leatrinanes12 | 2025-06-30