a_n = S_n - S_{n-1} = n(n+1)(n+2)/6 - (n-1)n(n+1)/6. Factor out n(n+1)/6 to get [n(n+1)/6] * [(n+2) - (n-1)] = [n(n+1)/6] * [3] = n(n+1)/2. This is the formula for the nth triangular number.

Group the terms in pairs: (1 - 2) + (3 - 4) + (5 - 6) + ... + (99 - 100). There are 100 terms, so there are 50 pairs. Each pair evaluates to -1. The sum is 50 * (-1) = -50.

This is a standard algebraic property. If a^2, b^2, c^2 are in AP, then adding ab+bc+ca to each term preserves the AP. Thus a^2+ab+bc+ca, b^2+ab+bc+ca, c^2+ab+bc+ca are in AP. Factoring gives (a+b)(a+c), (b+c)(a+b), (c+a)(c+b) are in AP. Dividing each term by (a+b)(b+c)(c+a) yields 1/(b+c), 1/(c+a), 1/(a+b) which are therefore in AP.

The relationship between AM (A), GM (G), and HM (H) is G^2 = A * H. We are given G = 8 and H = 6.4. Substituting these values: 8^2 = A * 6.4 => 64 = A * 6.4 => A = 64 / 6.4 = 10.

The differences between consecutive terms are 1, 2, 3, 4, ... which form an AP. The nth term of such a sequence is given by a_n = a_1 + Sum(first n-1 differences). a_10 = 1 + (1 + 2 + 3 + ... + 9) = 1 + 9(10)/2 = 1 + 45 = 46.

S = a/(1-r) = 3 => a = 3(1-r). The sum of first two terms is a + ar = a(1+r) = 8/3. Substitute a = 3(1-r): 3(1-r)(1+r) = 8/3 => 3(1-r^2) = 8/3 => 1-r^2 = 8/9 => r^2 = 1/9 => r = ±1/3. If r = 1/3, a = 3(1-1/3) = 2. If r = -1/3, a = 3(1+1/3) = 4. The first term can be 2 or 4.

Let the numbers be a, ar, ar^2. Given a(1+r+r^2) = 21 and a^2(1+r^2+r^4) = 189. Note that 1+r^2+r^4 = (1+r+r^2)(1-r+r^2). Dividing the second equation by the square of the first: a^2(1+r^2+r^4) / [a^2(1+r+r^2)^2] = 189 / 441. This simplifies to (1-r+r^2)/(1+r+r^2) = 3/7. Cross-multiplying gives 7-7r+7r^2 = 3+3r+3r^2 => 4r^2 - 10r + 4 = 0 => 2r^2 - 5r + 2 = 0. Factoring gives r=2 or r=1/2. If r=2, a(1+2+4) = 21 => 7a=21 => a=3. The numbers are 3, 6, 12.

Sum of all natural numbers from 1 to 100 = 100(101)/2 = 5050. The numbers divisible by 5 form an AP: 5, 10, ..., 100. The number of terms is 20. The sum of this AP is (20/2)(5 + 100) = 10 * 105 = 1050. The sum of numbers not divisible by 5 is the total sum minus the sum of multiples of 5: 5050 - 1050 = 4000.

The first AP consists of multiples of 2, its last term is 2*100=200. The second AP consists of multiples of 3, its last term is 3*80=240. The identical terms are numbers that are multiples of both 2 and 3, meaning they are multiples of LCM(2,3) = 6. We need to find the number of multiples of 6 that are less than or equal to the smaller maximum value (200). 200 / 6 = 33.33... So, there are 33 identical terms.

The nth term is a_n = S_n - S_{n-1}. So a_15 = S_15 - S_14. S_15 = 3(15^2) + 2(15) = 3(225) + 30 = 675 + 30 = 705. S_14 = 3(14^2) + 2(14) = 3(196) + 28 = 588 + 28 = 616. Therefore, a_15 = 705 - 616 = 89. Alternatively, a_n = (3n^2+2n) - (3(n-1)^2+2(n-1)) = 6n - 1. For n=15, 6(15) - 1 = 89.