Number Series

In an arithmetic progression, a_n = a + (n-1)d. Given a_3 = a + 2d = -9 and a_8 = a + 7d = 6, subtracting these gives 5d = 15, so d = 3. Then a + 2(3) = -9, so a = -15. Setting a_n = -15 + (n-1)3 = 0 gives 3(n-1) = 15, so n-1 = 5, n = 6.

The sequence is an arithmetic progression with first term a = 180 and common difference d = 180. The 20th term is a + (20-1)d = 180 + 19 * 180 = 20 * 180 = 3600.

Sequence 1: a = 53, d = 2. n-th term = 53 + (n-1)2 = 51 + 2n. Sequence 2: a = 1, d = 4. n-th term = 1 + (n-1)4 = 4n - 3. Set them equal: 51 + 2n = 4n - 3. 54 = 2n, so n = 27.

The sequence is an arithmetic progression with first term a = 3 and common difference d = 12. The 58th term is a + 57d = 3 + 57 * 12 = 3 + 684 = 687. We want the term that is 132 more than this, so 687 + 132 = 819. Setting a + (n-1)d = 819, we get 3 + (n-1)12 = 819, so (n-1)12 = 816, n-1 = 68, n = 69.

Sequence: 15, 25, 38, 54. Differences: 10, 13, 16. This is a quadratic sequence. Second difference is 3. General term T(n) = an^2 + bn + c. 2a = 3 => a = 1.5. 3a + b = 10 => 4.5 + b = 10 => b = 5.5. a + b + c = 15 => 1.5 + 5.5 + c = 15 => c = 8. T(n) = 1.5n^2 + 5.5n + 8. T(18) = 1.5(324) + 5.5(18) + 8 = 486 + 99 + 8 = 593. T(32) = 1.5(1024) + 5.5(32) + 8 = 1536 + 176 + 8 = 1720. Sum = 593 + 1720 = 2313.

The differences between consecutive terms are: 43-17=26, 81-43=38, 131-81=50. The differences increase by 12 each time (26, 38, 50). The next difference should be 50+12=62. 131+62=193.

The sequence is 3, 6, 15, 42, 123. Differences are 3, 9, 27, 81. This is a geometric progression with a=3, r=3. The sum of 8 terms of the differences is 3 * (3^8 - 1) / (3 - 1) = 3 * 6560 / 2 = 9840.

The pattern is based on adding consecutive Fibonacci-like numbers or specific increments. The differences between terms are: 133 - 83 = 50, 83 - 50 = 33, 50 - 33 = 17. Following this pattern (33 - 17 = 16), the missing term is 16 + 1 = 17.

The series follows a pattern of subtracting multiples of 5: 125 - 45 = 80, 80 - 35 = 45, 45 - 25 = 20, 20 - 15 = 5.

The sequence is 2, 4, 6, 8, which is 2*1, 2*2, 2*3, 2*4. The n-th term is 2n.

This is an arithmetic progression with a = 4 and d = 4. The 11th term is 4 + 10*4 = 44. The average of an AP is (first + last) / 2 = (4 + 44) / 2 = 24.

The sequence 0, 13, 26, 39 is an arithmetic progression with a first term a = 0 and a common difference d = 13. The formula for the n-th term is a + (n - 1)d, which simplifies to 0 + (n - 1)13 = 13n - 13.

This is an arithmetic progression with a = -7 and d = 4. The 58th term = a + (n-1)d = -7 + (57 * 4) = -7 + 228 = 221.