SOLUTION: the sum of the first and second terms of a geometric progression is 108 and the sum of the third and fourth is 12. find the two possible values of the common ratio and the correspo

Algebra ->  Triangles -> SOLUTION: the sum of the first and second terms of a geometric progression is 108 and the sum of the third and fourth is 12. find the two possible values of the common ratio and the correspo      Log On


   

Question 1171167: the sum of the first and second terms of a geometric progression is 108 and the sum of the third and fourth is 12. find the two possible values of the common ratio and the corresponding values of the first term
Found 3 solutions by MathLover1, greenestamps, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

a%5B1%5D%2Ba%5B2%5D=108
a%5B2%5D=a%5B1%5D%2Ar%5E%282-1%29
a%5B2%5D=a%5B1%5D%2Ar

then

a%5B1%5D%2Ba%5B1%5D%2Ar=108
a%5B1%5D%281%2Br%29=108
%281%2Br%29=108%2Fa%5B1%5D
r=108%2Fa%5B1%5D-1 ..............eq.1


and the sum of the third and fourth is 12

a%5B3%5D%2Ba%5B4%5D=12
since
a%5B3%5D=a%5B1%5D%2Ar%5E2
a%5B4%5D=a%5B1%5D%2Ar%5E3

we have
a%5B1%5D%2Ar%5E2%2Ba%5B1%5D%2Ar%5E3=12
a%5B1%5D%28r%5E2%2Br%5E3%29=12
r%5E2%2Br%5E3=12%2Fa%5B1%5D
r%28r%2Br%5E2%29=12%2Fa%5B1%5D........substitute r from eq.1
%28108%2Fa%5B1%5D-1%29%28108%2Fa%5B1%5D-1%2B%28108%2Fa%5B1%5D-1%29%5E2%29=12%2Fa%5B1%5D
%28108%2Fa%5B1%5D-1%29%28108+%28108+-+a%5B1%5D%29%29%2Fa%5B1%5D%5E2=12%2Fa%5B1%5D
%28108+%28a%5B1%5D+-+108%29%5E2%29%2Fa%5B1%5D%5E3+-12%2Fa%5B1%5D=0
:
:
comes to factored numerator:
%28a%5B1%5D+-+162%29+%28a%5B1%5D+-+81%29+=+0++

solutions:
a%5B1%5D+=+162
or
a%5B1%5D+=+81

since given a%5B1%5D%2Ba%5B2%5D=108, if a%5B1%5D+=+162
162%2Ba%5B2%5D=108
a%5B2%5D=108-162
a%5B2%5D=-54

r=-54%2F162
r=-1%2F3

now find third and fourth term

then
a%5B2%5D=108-81
a%5B2%5D=27

and
r=27%2F81
r=1%2F3

now find third and fourth term
a%5B3%5D=a%5B1%5D%2Ar%5E2=81%281%2F3%5E2%29=81%281%2F9%29=9
a%5B4%5D=a%5B1%5D%2Ar%5E3=81%281%2F3%5E3%29=81%281%2F27%29=3

so, the two possible values of the common ratio and the corresponding values of the first term are:
1. a%5B1%5D+=+162 and r=-1%2F3
2. a%5B1%5D+=+81 and r=1%2F3


Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


The solution from the other tutor is a good example of the fact that there are practically endless ways to solve any given problem....

Here is a slightly simpler method.

(1) a%2Bar+=+108 the sum of the 1st and 2nd terms is 108

(2) ar%5E2%2Bar%5E3+=+9 the sum of the 3rd and 4th terms is 12

Factor out a common factor on the left in (2); then substitute (1):

ar%5E2%2Bar%5E3+=+r%5E2%28a%2Bar%29+=+108r%5E2+=+12
r%5E2+=+12%2F108+=+1%2F9
r+=+1%2F3 or r+=+-1%2F3

If r=1/3 then

a%2Ba%281%2F3%29+=+108
%284%2F3%29a+=+108
a+=+108%283%2F4%29+=+81

ANSWER #1: a=81; r=1/3. The terms are 81, 27, 9, 3.

CHECK: 81+27 = 108; 9+3 = 12.

If r=-1/3 then

a%2Ba%28-1%2F3%29+=+108
%282%2F3%29a+=+108
a+=+108%283%2F2%29+=+162

ANSWER #2: a=162; r=-1/3. The terms are 162, -54, 18, -6.

CHECK: 162+(-54) = 108; 18+(-6) = 12.

Answer by ikleyn(52878) About Me  (Show Source):
You can put this solution on YOUR website!
.

The post by @MathLover1 is a good example of  EXTREMELY  BAD  approach and  EXTREMELY  BAD  presentation.


It is an example of  WHAT  a  SOLUTION  SHOULD  NOT  be.


/\/\/\/\/\/\/\/


Dear @MathLover1

teaching  Math and presenting solutions to  Math problems should be easy and elegant - then  (and only then)
the students will love  Math.


Otherwise,  they will  HATE  it . . .

If it is difficult for you to provide such teaching uniformly in many areas,
consider to focus in that areas where you are an expert.


Do not try to embrace everything - it is wrong way and wrong strategy . . .


Leave it to true experts . . .