The Arithmetic Progression Problem
A sequence of numbers is defined as follows: the first term is 5, and each subsequent term is obtained by adding 3 to the previous term. The sequence continues for 20 terms. A student is asked to find the sum of all terms in the sequence. Additionally, the student is told that the sum of the first n terms is given by the formula S_n = n/2 * (2a + (n-1)d), where a is the first term and d is the common difference.
1. Calculate the 20th term of the sequence and the sum of the first 20 terms using the given formula.
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The sequence is arithmetic with a = 5, d = 3. The nth term is a_n = a + (n-1)d = 5 + 193 = 5 + 57 = 62. So the 20th term is 62. Sum S_20 = 20/2 * (25 + 19*3) = 10 * (10 + 57) = 10 * 67 = 670.
2. If the sum of the first n terms is 1,000, find the value of n. (Assume n is an integer.)
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S_n = n/2 * (25 + (n-1)3) = n/2 * (10 + 3n - 3) = n/2 * (3n + 7). Set equal to 1000: n(3n+7)/2 = 1000 → 3n² + 7n = 2000 → 3n² + 7n - 2000 = 0. Solve using quadratic formula: n = [-7 ± √(49 + 24000)]/6 = [-7 ± √24049]/6. √24049 ≈ 155.08 (since 155²=24025, 156²=24336). So n ≈ (-7 + 155.08)/6 = 148.08/6 ≈ 24.68. Since n must be integer, check n=24: S_24 = 24/2(324+7)=12*(72+7)=1279=948. n=25: S_25=25/2(75+7)=12.5*82=1025. So no integer n gives exactly 1000. The closest is n=24 (sum 948) or n=25 (sum 1025). Thus, no integer solution.