SOLUTION: My problem is to find the indicated term for the geometric series described : Sn=33 an=48 r=-2 , find a1. I am trying to use the formula SN= a1-a1 r(n)/ 1-r However, i am

Question 843435: My problem is to find the indicated term for the geometric series described :
Sn=33 an=48 r=-2 , find a1.
I am trying to use the formula SN= a1-a1 r(n)/ 1-r
However, i am missing 2 variables both a1 and "n". I have tried to solve first using an=a1 r (n-1) but again, i am missing 2 variables (a1 and n).
Please tell me how to solve for a1. Thanks
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
Sn = 33
An = 48
R = 2

The formula for the Last term of a geometric series is An = A1 * (R)^(n-1)

The formula for the sum of a geometric series is Sn = A1 * (1 - r^n) / (1 - r)

I was actually able to find A1 and n but it wasn't easy.

I got n = .60768258

Once I got n, I was able to find A1.

I got A1 = 63

Your answer is A1 = 63.

I tried to solve it by formula but wound up in a dead end because I was getting logs of a negative number which is not allowed.

Solving simultaneous equations by graphing is a legitimate solution technique.
I'm not sure if I should have been able to solve it by formula, but I couldn't do it, so graphing was the only other way I knew how to find it.

What I did was solve for A1 in both equations and then set those equations equal to each other.

I then subtracted the second equation from the first and graphed it.

My solution was where the graph crosses the x-axis.

The two equations that I subtracted from each other are equal at that point.

Once I found the value of n, I was able to substitute for n in either equation to get A1.

Here's the details of what I did.

The first equation is:

An = A1 * (2^(n-1))
Solve for A1 to get A1 = An / (2^(n-1))
Replace An with 48 to get A1 = 48 / (2^(n-1))

The second equation is:

Sn = A1 * (1 - 2^n) / (1 - 2)
Simplify to get:
Sn = A1 * (1 - 2^n) / (-1)
Simplify further to get:
Sn = (-A1) * (1 - 2^n)
Replace Sn with 33 to get:
33 = (-A1) * (1 - 2^n)
divide both sides of this equation by (1 - 2^n) to get:
33 / (1 - 2^n) = -A1
multiply both sides of this equation by -1 to get:
-33 / (1 - 2^n) = A1
commute this equation (flip sides) to get:
A1 = -33 / (1 - 2^n)

The equations that I got for A1 are:
A1 = 48 / 2^(n-1)
and:
A1 = -33 / (1 - 2^n)

Since both expressions on the right side of these equations are equal to A1, I set the expressions on the right side of each equation equal to each other to get:

48 / 2^(n-1) = -33 / (1 - 2^n)

I tried to solve this using logs but wound up with logs of a negative number which isn't allowed.

That's when I resorted to graphing.

I added the right side of the equation to both sides of the equation to get:

48 / 2^(n-1) + 33 / (1 - 2^n) = 0

I then set this equation equal to y and graphed it.

The equation that was graphed is:

y = 48 / (2^(x-1)) + 33 / (1 - 2^x)

I had to change n to x in order for the graphing software to work.

the graph of that equation is shown below:

You can't really get the value from the graph.
I used the TI-84 which tells you what the zero crossing point is.
I then use the same TI-84 to find A1 which it told me was 63.

The value of A1 = 63 was generated using the internally stored value of n rather than the rounded value of n that is shown above.

The actual zero crossing shown on my TI-84 was x = .60768258.

That's the value of n. It doesn't look rounded but it is. The internally stored number carries it out to more decimal places.

Use that value and you get A1 = something close to 63, but not 63.
When I used the internally stored number from the TI-84 I got 63 exactly.

Anyway, the problem is solved.
A1 = 63

The method to find the solution was graphing of the equations that were used to solve for A1 in terms of n.

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