2 questions about terms and sequences?

Question

Okay, so I know how to calculate sequences and terms, but I can't figure out how to do these two questions:

1) An arithmetic sequence and a geometric sequence, each of only 3 terms, have the same middle term. The sum of the arithmetic sequence is 18, and the sum of the geometric sequence is 21. What are the terms of the geometric sequence?

2) A formula for the sum of the first n terms of an arithmetic series is Sn = 0.5n - n^2. Determine the 11th term of the series.

Answer 1

Q1.
First, find the middle term in the arithmetic sequence:
a+2a+3a=18
6a=18
a=3
so 2a=6. Yay.
Now set up the geometric sequence:
c + cn^1 + cn^2 = 21
but we know the middle term,
c + 6 + cn^2 = 21
c + cn^2 = 15
Factor: c(1+n^2) = 15 = 3*5
so c = 3, n = 2
now we get 3, 3*2, 3*2*2
or 3, 6, 12

Q2.
Sn = S(n-1) + an, where an is the nth term in the sequence, right?
Just solve for an = Sn - S(n-1) and put in n = 11.
0.5*11 - 11^2 - 0.5*10 + 10^2 = -20.5

Answer 2

1. Since the middle term of a 3-term arithmetic sequence is also its mean, here the middle term is 6.

The middle term of a 3-term geometric sequence is the square root of the product of the previous term and the following term. So The geometric terms are a, 6, b where ab = 36 and a + 6 + b = 21

Now a + b must be 21-6 = 15

a = 15 - b

Substitute in the equation ab=36 to get (15-b)b=36

15b - b^2 = 36

b^2 - 15b + 36 = 0

(b - 12)(b - 3) = 0

etc


2. The sum of the first term (which would actually BE the first term) is 0.5(1) - 1^2 which is -1/2

The sum of the first 2 terms is 0.5(2) - 2^2 which is -3, so the second term must have been - 2 1/2 (negative two and a half). This is 5 times the first term so the common ratio is 5. Plug this in to the formula for the nth term.

Answer 3

1) Lets show the first 3 terms of arithmetic as ,a,a+b,a+2b.
So, a+a+b+a+2b=18, or 3a+3b=18, so a+b=6.
Now, lets show 3 terms of geometric as, c,cd,cd^2.
So, c+cd+cd^2=21.
Now, since a+b=6=cd, then c+6+.6d=21.
So, since cd=6, or c=6/d, then (6/d)+6+6d=21, or (6/d)+6d=15, or 6+6d^2=15d, or 6d^2-15d+6=0. This can factor to (3d-6 )(2d-1)=0. So, 1 solution is 3d-6=0, or d=2.
So, 1 solution is 3,6,12.
And, i guess another could be 12,6,3.

2)The 11th term is ((.5*11)-11^2)-((.5*10)-10^2) = -115.5-(-95)=-20.5.

Answer 4

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