Arithmetical ProgressionDate: 7/7/96 at 1:55:28 From: Wong Cheong Siong Subject: Arithmetical Progression Dear Dr. Math, An arithmetical progression has a common difference of 1/1/2. The sum of the first n terms is 365 and the sum of the first 2n terms is 1330. Calculate the value of n and the first term. Best Regards, Wong Cheong Siong Date: 7/7/96 at 16:34:41 From: Doctor Anthony Subject: Re: Arithmetical Progression a = first term, d= 1.5 (I think, because you wrote it 1/1/2 which could equal 2) Sn = (n/2)(2a+(n-1)d) First situation (n/2)(2a + (n-1)*1.5) = 365 Second situation (2n/2)(2a + (2n-1)*1.5) = 1330 These can be written n(2a + 1.5(n-1)) = 730 n(2a + 1.5(2n-1)) = 1330 Subtract top equation from the second equation. This elimimates 2an term. We get 1.5n(2n-1 - n+1) = 600 1.5n(n) = 600 n^2 = 400 n = 20 To get first term we can write 20(2a + 1.5*19) = 730 4a + 3*19 = 73 4a = 16 a = 4 Check if this works in second situation. 20(8 + 1.5*39) = 1330 which checks. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 7/9/96 at 3:45:12 From: Wong Cheong Siong Subject: Re: Arithmetical Progression Dear Dr. Math, I would like to thank you for answering my question. However, my teacher told us that a question can be done in very few steps, but he will not tell us how. So, can I ask you to show me the fastest possible way to solve this question: The sum of n terms of a certain series is 3n^2 + 10n for all values of n. Find the nth term and show that the series is an arithmetical progression. Best Regards, Wong Cheong Siong Date: 7/9/96 at 7:28:55 From: Doctor Anthony Subject: Re: Arithmetical Progression To show whether it is an AP or a GP you can find the first three terms using the formula for Sn = 3n^2 + 10n S1 = 3 + 10 = 13 a1 = 13 S2 = 12 + 20 = 32 a2 = S2-S1 = 19 S3 = 27 + 30 = 57 a3 = S3-S2 = 25 The series has a = 13 and there is a common difference d = 6 The nth term is given by a + (n-1)d = 13 + (n-1)*6 = 6n + 7 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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