Sequences and Series

The series 1, 8, 15 is an arithmetic progression with common difference 7. Sum of n terms = n/2 × (2a + (n-1)d). For n=10: 10/2 × (2×1 + 9×7) = 5 × (2 + 63) = 5 × 65 = 325.

Using the arithmetic series formulas: nth term aₙ = a₁ + (n-1)d and sum Sₙ = n(a₁ + aₙ)/2. Given a₃ = 114, aₙ = -27, and S = 2325. From a₃ = a₁ + 2d = 114, we get a₁ = 114 - 2d. From S = n(a₁ + aₙ)/2 = n((114-2d) + (-27))/2 = n(87 - 2d)/2 = 2325. From aₙ = a₁ + (n-1)d: -27 = (114-2d) + (n-1)d = 114 - 2d + nd - d = 114 + nd - 3d. Solving gives d = -3, n = 60, a₁ = 120. The last three terms are: 58th = a₁ + 57d = 120 + 57(-3) = -51, 59th = 120 + 58(-3) = -54, 60th = 120 + 59(-3) = -57. Wait - this doesn't match option D. Let me recalculate: -27 = a₁ + (n-1)(-3), so a₁ = -27 + 3(n-1). Also a₁ = 114 - 2(-3) = 120. So 120 = -27 + 3n - 3, so 3n = 150, n = 50. Last three: 48th = 120 + 47(-3) = -21, 49th = 120 + 48(-3) = -24, 50th = 120 + 49(-3) = -27. This matches option D.

The series is an arithmetico-geometric progression. The sum of 1 + 2x + 3x^2 + ... is given by the formula 1/(1-x)^2 for |x| < 1. Substituting x = 0.5, we get 1/(1-0.5)^2 = 1/(0.5)^2 = 1/0.25 = 4.

Let terms be a, ar, ar^2, ar^3. a(1+r)=12 and ar^2(1+r)=48. Dividing gives r^2=4, so r=2 or -2. Since terms alternate signs, r must be -2. Then a(1-2)=12 => -a=12 => a=-12.

Let AP1 be: a, a+d, a+2d, ... and AP2 be: b, b+e, b+2e, ... The product sequence is: ab, (a+d)(b+e), (a+2d)(b+2e), ... Term 1: 192 = ab, Term 2: 360 = (a+d)(b+e) = ab + ae + bd + de, Term 3: 576 = (a+2d)(b+2e). From 192 to 360, difference is 168. From 360 to 576, difference is 216. The second difference is 216-168=48. For product of two APs, the 8th term follows the pattern. Through calculation: if we factor 192 = 12×16, 360 = 15×24, 576 = 18×32, then AP1: 12, 15, 18, 21, 24, 27, 30, 33 and AP2: 16, 24, 32, 40, 48, 56, 64, 72. Term 8 = 33 × 72 = 2376.

The given series converges to 1.5.

Divide each term by 2. So, 32 ÷ 2 = 16 and 16 ÷ 2 = 8 So, 16 and 8 are the next two terms.

3, 5, 7, 9, ... 3 + 2 = 5 5 + 2 = 7 7 + 2 = 9 Rule: Add 2 to get the next term. So, the next term = 9 + 2 = 11

128, 64, 32, 16, ... Pattern: Divide each term by 2 to get the next term. So, the next two terms are 8 and 4.

In the series: 2, 6, 12, 20, ..., each number is written as the sum of its square and the number itself. 2 = 1² + 1 6 = 2² + 2 12 = 3² + 35² + 5 = 25 + 5 = 30 6² + 6 = 36 + 6 = 42 Hence, their sum would be 72.  

For a GP with second term ar = 24 and fifth term ar^4 = 81, we divide to find r^3 = 81/24 = 27/8, so r = 3/2. Then a = 24/r = 24/(3/2) = 16. The series is 16, 16×(3/2), 16×(3/2)^2, ... = 16, 24, 36, 54, 81,...

It is the correct option. First term = 2 Second term = 1/4*(23) = 2 Third term = 1/4*(23) = 2 Fourth term = 1/4*(23) = 2 So, each term of the given sequence is 2 only. So, option (2) is correct.