Arithmetical Progression

Date: 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/