The sum of the first and second terms of a geometric progression is -5, and the sum of the fourth and fifth terms is 40. Find the first term and the common ratio.

8. (i) Step 1: Write down the given information in terms of the first term a and common ratio r of a geometric progression. The n-th term of a geometric progression is a_n = ar^n-1. The sum of the first and second terms is -5: a_1 + a_2 = -5 a + ar = -5 a(1+r) = -5 (Equation 1) The sum of the fourth and fifth terms is 40: a_4 + a_5 = 40 ar^3 + ar^4 = 40 ar^3(1+r) = 40 (Equation 2) Step 2: Solve the system of equations for r. Divide Equation 2 by Equation 1: (ar^3(1+r))/(a(1+r)) = (40)/(-5) Assuming a ≠ 0 and 1+r ≠ 0: r^3 = -8 r = [3]-8 r = -2 Step 3: Solve for a. Substitute r=-2 into Equation 1: a(1+(-2)) = -5 a(1-2) = -5 a(-1) = -5 a = 5 The first term is a=5 and the common ratio is r=-2. First term = 5 Common ratio = -2 8. (ii) Step 1: Write the general term of the binomial expansion. The general term T_k+1 in the expansion of (A+B)^n is given by T_k+1 = nk A^n-k B^k. For (x + (2)/(x^2))^15, we have A=x, B=(2)/(x^2), and n=15. T_k+1 = 15k (x)^15-k ((2)/(x^2))^k Step 2: Simplify the general term to combine the powers of x. T_k+1 = 15k x^15-k (2^k)/((x^2)^k) T_k+1 = 15k x^15-k (2^k)/(x^2k) T_k+1 = 15k 2^k x^15-k-2k T_k+1 = 15k 2^k x^15-3k Step 3: Find the value of k for the term independent of x. For the term to be independent of x, the exponent of x must be 0. 15 - 3k = 0 3k = 15 k = 5 Step 4: Substitute k=5 into the general term to find the independent term. The term independent of x is T_5+1 = T_6: T_6 = 155 2^5 x^15-3(5) T_6 = 155 2^5 x^0 T_6 = 155 2^5 Calculate 155: 155 = (15!)/(5!(15-5)!) = (15!)/(5!10!) = (15 × 14 × 13 × 12 × 11)/(5 × 4 × 3 × 2 × 1) 155 = (360360)/(120) = 3003 Calculate 2^5: 2^5 = 32 Multiply the values: T_6 = 3003 × 32 T_6 = 96096 The term independent of x is 96096. Drop the next question.