# Arithmetic Sequence Series Problem

1. Nov 16, 2004

### nobb

Hi. Please explain to me how to do these three problems:

1. The terms x+3, 3x-1, and 7x-2 are consecutive terms in an arithmetic sequence. Find x.

2. The sum of the first twenty-eight terms of an arithmetic series if the sum of the first twelve terms is -72 and the sum of the first twenty-four terms is 144.

3. Find a, d, and tn for this sequence: t4= -9 , t15 = -31

2. Nov 17, 2004

### HallsofIvy

The terms in an arithmetic sequence always vary by a constant:
a+ r, a+ 2r, a+ 3r, etc so that subtracting and consecutive two terms gives r.

1. Here we must have (3x-1)- (x+1)= r= (7x-2)- (3x-1). Ignore the r and solve for x.

2. With an arithmetic sequence you can find the "average" of all numbers by averaging the first and last terms of the sequence. The sum of n terms is just n times that average.

In particular the average of the first 12 terms is (a1+ a12)/2 so the sum of the first 12 terms is 12(a1+ a12)/2= 6(a1+ a12)= -72.

Similarly, the sum of the first 24 terms is 12(a1+ a24)= 144.

Of course, if we take a1 as the first term and d as the common difference, a12= a1+ 11d and a24= a1+23d.
Putting those into the two equations for the sums gives you 2 equations in the two unkowns, d and a1. Once you know those, you can calculate a28= a1+ 27d and the sum is 28(a1+ a28[/sub)/2= 14(a1+ a28).

a1= a and ann= a+ (n-1)d so t4 (what I have called a4= a+ 3d= -9 and t15= a+ 14d= -31. Solve those two equations for a and d and then tn= a+ (n-1)d.