Number Series

Sequence: 20, 19.25, 18.5, 17.75... Arithmetic progression with a = 20, d = -0.75. We want a + (n-1)d < 0. 20 + (n-1)(-0.75) < 0. 20 < (n-1)(0.75). 26.66 < n-1. n > 27.66. The first integer n is 28.

The sequence is a geometric progression with the first term a = 1/2 and common ratio r = 1/3. The next term is (1/54) * (1/3) = 1/162.

This is an arithmetic progression with a common difference of 3. The next term is 10 + 3 = 13.

The sequence is a geometric progression with a common ratio of 2. The next term is 24 * 2 = 48.

The sequence 2, 5, 8, 11 is an arithmetic progression with a common difference of 3. The next term is 11 + 3 = 14.

The sequence is an arithmetic progression with a common difference of -3 (-13 - (-10) = -3). The next term is -19 - 3 = -22.

The sequence is a decreasing arithmetic progression with a common difference of -1.

This is an arithmetic progression with first term a = m-1 and common difference d = -2. The n-th term = a + (n-1)d = (m-1) + (n-1)(-2) = m - 1 - 2n + 2 = m - 2n + 1.

Sequence is 3, 7, 11, 15... This is an AP with a=3, d=4. nth term = a + (n-1)d = 3 + (n-1)4 = 4n - 1. 9th term = 4(9) - 1 = 35.

Sum of arithmetic series = (n/2)[2a + (n-1)d]. Here a=18, d=-2. Sum = (n/2)[36 + (n-1)(-2)] = (n/2)[36 - 2n + 2] = (n/2)[38 - 2n] = n(19-n). For sum to be zero, n=19.

The sum of n terms where the n-th term is n(n+1) = n^2 + n is the sum of the first n squares plus the sum of the first n integers. This equals [n(n+1)(2n+1)/6] + [n(n+1)/2], which simplifies to n(n+1)(n+2)/3.

The sequence is an arithmetic progression with first term 11 and common difference 9. The general term is a_n = 11 + (n-1)9 = 9n + 2. Checking the options: 326 = 9n + 2 => 324 = 9n => n = 36. 173 = 9n + 2 => 171 = 9n => n = 19. 388 = 9n + 2 => 386 = 9n (not divisible).

This is an arithmetic progression with a = 9, d = 3, l = 66. Formula: l = a + (n-1)d. 66 = 9 + (n-1)3. 57 = (n-1)3. 19 = n-1, so n = 20.

This is an arithmetic progression with a=3 and d=4. The 17th term is a + (n-1)d = 3 + (16 * 4) = 3 + 64 = 67.

The nth term is given by a_n = 3n - 1. For n=1, a_1 = 3(1)-1 = 2. For n=2, a_2 = 3(2)-1 = 5. For n=3, a_3 = 3(3)-1 = 8. The series is 2, 5, 8, ...