Arithmetic Sequence Series Problem

[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

 

[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+ a<sub>28[/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.</sub>

 

[wais]

1. To find x, we can use the formula for the nth term of an arithmetic sequence, which is an = a1 + (n-1)d, where a1 is the first term and d is the common difference. In this case, we have a1 = x+3, a2 = 3x-1, and a3 = 7x-2. Plugging these values into the formula, we get:

a1 = x+3
a2 = (x+3) + d = 3x-1
a3 = (x+3) + 2d = 7x-2

Solving the first two equations for d and x, we get d = 2 and x = 5. Therefore, the terms are 8, 9, and 17 in an arithmetic sequence.

2. To find the sum of the first twenty-eight terms, we can use the sum formula for an arithmetic series, which is Sn = n/2(a1 + an), where n is the number of terms, a1 is the first term, and an is the last term. We are given the sum of the first twelve and first twenty-four terms, so we can set up the following equations:

12/2(a1 + a12) = -72
24/2(a1 + a24) = 144

Solving for a1 and a24, we get a1 = -9 and a24 = 15. Then, plugging these values into the sum formula, we get:

28/2(-9 + a28) = -72
a28 = -27

Therefore, the sum of the first twenty-eight terms is -27.

3. To find a, d, and tn, we can use the same formula for the nth term of an arithmetic sequence. We are given two terms, t4 = -9 and t15 = -31, so we can set up the following equations:

t4 = a + 3d = -9
t15 = a + 14d = -31

Solving for a and d, we get a = -3 and d = -2. Then, using the formula for tn, we can find the 4th and 15th terms:

t4 = -3 + 3(-2) = -9
t15 = -3 + 14(-2) = -31