SOLUTION: Find the 12th term of a Geometric Sequence whose 3rd term is 432 and the 5th term is 15552

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Question 1197284: Find the 12th term of a Geometric Sequence whose 3rd term is 432 and the 5th term is 15552
Found 3 solutions by MathLover1, greenestamps, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!


general term is:
a%5B+n%5D=+a+%5B1%5D%2A+r%5E%28+n+-1%29

given:
3rd term is a%5B3%5D=432
5th term is a%5B5%5D=15552


find a%5B1%5D and r

432=+a+%5B1%5D%2A+r%5E%28+3+-1%29
432=+a+%5B1%5D%2A+r%5E2...........solve for a%5B1%5D
a+%5B1%5D=432%2Fr%5E2..............eq.1


15552=+a+%5B1%5D%2A+r%5E%28+5+-1%29
15552=+a+%5B1%5D%2A+r%5E4...........solve for a%5B1%5D
a+%5B1%5D=15552%2Fr%5E4..............eq.2


from eq.1 and eq.2 we have

432%2Fr%5E2=15552%2Fr%5E4........cross multiply
432r%5E4=15552r%5E2..........simplify
432r%5E2=15552
r%5E2=15552%2F432
r%5E2=36
r=6

go to

a+%5B1%5D=432%2Fr%5E2..............eq.1, substitute r
a+%5B1%5D=432%2F6%5E2
a+%5B1%5D=432%2F36
a+%5B1%5D=12

your nth term formula is:

a%5B+n%5D=+12%2A+6%5E%28+n+-1%29

then, the 12th term will be

a%5B+12%5D=+12%2A+6%5E%28+12+-1%29

a%5B+12%5D=+12%2A+6%5E11

a%5B+12%5D=+4353564672




Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


The 5th term is the 3rd term, multiplied by the common ratio twice. Divide the given 5th term by the given 3rd term and take the square root of the result to find the common ratio.

Then to find the 12th term, multiply the given 5th term by the common ratio 7 times.

Note that the given information only gives us the square of the common ratio, so that common ratio might be either positive or negative. Since the given 5th term is positive and you are multiplying by the common ratio an odd number of times, the 12th term could be either positive or negative.


Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find the 12th term of a Geometric Sequence whose 3rd term is 432 and the 5th term is 15552.
~~~~~~~~~~~~~~~~


            @MathLover1 wrote many symbols, but failed to solve the problem in a right way:
            she missed one of the two solutions.

            So I came to bring a correct solution and correct answer.


Since the 3rd and the 5th terms of the geometric progression are given, we write

    a%5B5%5D = a%5B3%5D%2Ar%5E2,

or

    15552 = 432%2Ar%5E2.


From this equation,  

    r%5E2 = 15552%2F432 = 36.


Hence,  r = sqrt%2836%29 = +/- 6.


Thus, there are TWO geometric progressions satisfying the given properties:
one progression with the common ratio r= 6 and the other progression with the common ratio r= -6.



For the first progression

    a%5B12%5D = a%5B5%5D%2Ar%5E%2812-5%29 = 15552%2A6%5E7 = 4353564672.



For the second progression

    a%5B12%5D = a%5B5%5D%2Ar%5E%2812-5%29 = 15552%2A%28-6%29%5E7 = -4353564672.



Thus there are two answers and two values for the 12-th term: 4353564672 and -4353564672.

Solved.

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On geometric progressions,  see introductory lessons
    - Geometric progressions
    - The proofs of the formulas for geometric progressions
    - Problems on geometric progressions
    - Word problems on geometric progressions
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic
"Geometric progressions".

Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.