This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
a) The first three consecutive terms of a geometric progression (GP) are 3^2x+1, 9^x, and 81. In a GP, the ratio of consecutive terms is constant. So, we can write: (9^x)/(3^2x+1) = (81)/(9^x) To solve for x, we can express all terms with a base of 3: 9 = 3^2 and 81 = 3^4. Substitute these into the equation: ((3^2)^x)/(3^2x+1) = (3^4)/((3^2)^x) 3^2x3^2x+1 = (3^4)/(3^2x) Using the exponent rule (a^m)/(a^n) = a^m-n: 3^2x - (2x+1) = 3^4 - 2x 3^-1 = 3^4 - 2x Since the bases are equal, the exponents must be equal: -1 = 4 - 2x 2x = 4 + 1 2x = 5 x = (5)/(2) x = 2.5 The value of x is 2.5. b) To find the common ratio (r) of the series, we can use the value of x found in part (a). The terms are 3^2x+1, 9^x, 81. Substitute x = 2.5: First term (T_1): 3^2(2.5)+1 = 3^5+1 = 3^6 = 729 Second term (T_2): 9^2.5 = (3^2)^2.5 = 3^5 = 243 Third term (T_3): 81 The common ratio r is (T_2)/(T_1) or (T_3)/(T_2). r = (243)/(729) = (1)/(3) Alternatively, r = (81)/(243) = (1)/(3) The common ratio of the series is (1)/(3). c) To calculate the sum of the first 4 terms (S_4) of the series, we use the formula for the sum of the first n terms of a GP: S_n = (a(1-r^n))/(1-r), where a is the first term and r is the common ratio. From parts (a) and (b): First term, a = 729 Common ratio, r = (1)/(3) Number of terms, n = 4 Substitute these values into the formula: S_4 = (729(1 - (1)/(3))^4)1 - (1)/(3) S_4 = (729(1 - 1)/(81))(2)/(3) S_4 = (729(81-1)/(81))(2)/(3) S_4 = (729(80)/(81))(2)/(3) S_4 = (9 × 80)/(2)3 S_4 = (720)/(2)3 S_4 = 720 × (3)/(2) S_4 = 360 × 3 S_4 = 1080 The sum of the first 4 terms of the series is 1080. d) We are given that the fifth and the seventh terms of the GP form the first two consecutive terms of an arithmetic sequence (AP). First, find the fifth term (T_5) and the seventh term (T_7) of the GP. The general term of a GP is T_n = ar^n-1. a = 729 and r = (1)/(3). Fifth term of GP (T_5): T_5 = ar^5-1 = ar^4 = 729 × ((1)/(3))^4 T_5 = 729 × (1)/(81) = 9 Seventh term of GP (T_7): T_7 = ar^7-1 = ar^6 = 729 × ((1)/(3))^6 T_7 = 729 × (1)/(729) = 1 These terms, T_5=9 and T_7=1, are the first two consecutive terms of an arithmetic sequence. Let the first term of the AP be A_1 = 9. Let the second term of the AP be A_2 = 1. Now, find the common difference (d) of this arithmetic sequence: d = A_2 - A_1 = 1 - 9 = -8 We need to calculate the sum of the first 20 terms (S_20) of this arithmetic sequence. The formula for the sum of the first n terms of an AP is S_n = (n)/(2)(2A_1 + (n-1)d). Here, n=20, A_1=9, and d=-8. Substitute these values into the formula: S_20 = (20)/(2)(2(9) + (20-1)(-8)) S_20 = 10(18 + (19)(-8)) S_20 = 10(18 - 152) S_20 = 10(-134) S_20 = -1340 The sum of the first 20 terms of the arithmetic sequence is -1340. That's 2 down. 3 left today — send the next one.