geometric series

SabziiKumari

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Jan 24, 2010
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find the five first terms in the geometric series which is the sum :

1st and 3rd term = 50
2nd and 4th term = 150

do solve this would you have to make simultanuous eqations.

1 .. a + ar2 = 50
2.. ar + ar3 = 150

sn4 = 200

what would i do next ?
thanks.
 
find the five first terms in the geometric series having the sums :

1st and 3rd term = 50
2nd and 4th term = 150

do solve this would you have to make simultanuous equations.

Let a = the first term, r = the common ratio and n = the number of terms.

The terme are then a, ar, ar^2, ar^3.......ar^(n-1).

a + ar^2 = 50
ar + ar^3 = 150
a(1 + r^2) = 50
a(r + r^3 = 150
Solving for "a" and equating the results yields r^3 - 3r^2 + r = 3 yielding r = 3 and a = 5.
Therefore, the series is 5, 15, 45, 135, 405,...
 
Solve the cubic equation from TchrWill, first.

r^3 - 3r^2 + r = 3

Once you know that r = 3, you could substitute 3 for r into one of the other equations, and solve for a.

a + a(3)^2 = 50
 
TchrWill said:
a(1 + r^2) = 50

a(r + r^3) = 150

Solving [each of the two equations above] for "a", and equating the results yields r^3 - 3r^2 + r = 3

The phrase "equating the results" means to set both expressions for "a" equal to one another.

50/(1 + r^2) = 150/(r + r^3)

This is a proportion. Cross-multiply, and simplify to obtain the cubic equation.
 
50/(1 + r^2) = 150/(r + r^3)

i dont get how you still got the cubic equations.
icant get head around it :(
 
When two ratios are equal, we call it a proportion.

A/B = C/D

We can eliminate the fractions in a proportion by "cross-multiplying". This means to multiply the numerator on the left by the denominator on the right AND to multiply the denominator on the left by the numerator on the right. The results are ALWAYS equal, as shown symbolically:

A*D = B*C

So, if we cross multiply the proportion at hand, we get:

50(r + r^3) = 150(1 + r^2)

Use the Distributive Property, to carry out the next multiplications.

(50)(r) + (50)(r^3) = (150)(1) + (150)(r^2)

50r + 50r^3 = 150 + 150r^2

Subtract 150r^2 from both sides, to get all r-terms on the lefthand side.

50r^3 - 150r^2 + 50r = 150

Note that all coefficients are divisible by 50; so, divide both sides by 50.

r^3 - 3r^2 = r = 3

Those last two steps could be done in either order. 8-)
 
SabziiKumari said:
once you get

r - 3r^2 + r^3

how do yu get that to equal 3 ?

Well, the previous equation was equal to 150, and we divided both sides by 50.

150/50 = 3

That's how r - 3r^2 + r^3 ends up equal to 3.

Have you taken any algebra classes?
 
yes i have thank you
your great help

i just thought you had to solve : r - 3r^2 + r^3 = 3


to get r on its own.
 
SabziiKumari said:
i just thought you had to solve : r - 3r^2 + r^3 = 3


to get r on its own.

Ah. Yes we do, but that is not what your last question is talking about.

You asked how we get r - 3r^2 + r^3 to equal 3.

What you're now trying to ask is actually, "How do we get the variable r to equal 3".

Subtract 3 from both sides, to get a zero on one side.

r^3 - 3r^2 + r - 3 = 0

One could factor the lefthand side, and then use the Zero Product Property (i.e., set each factor equal to zero).

One could use the Rational Roots Theorem to find possible solutions, followed by testing them to see which ones work.

I'm now thinking that you need serious algebra review, before attempting exercises involving geometric sequences. :idea:
 
thank you
i re read through all the note you had written and i do understand it so much better
i just skim read all the information.
next time will have to take better care.

thank you.
 
SabziiKumari said:
i do not understand how you a=5 and r =3

thnks

a + ar[sup:1s863y18]2[/sup:1s863y18] = 50 ? a(1+r[sup:1s863y18]2[/sup:1s863y18]) = 50 ...................................(1)

ar + ar[sup:1s863y18]3[/sup:1s863y18] = 150 ? ar(1+r[sup:1s863y18]2[/sup:1s863y18]) = 150................................(2)

divide (2) by (1) - you get

[ar(1+r[sup:1s863y18]2[/sup:1s863y18])]/[a(1+r[sup:1s863y18]2[/sup:1s863y18])] = 150/50

r = 3

using this value in (1)

a(1+3[sup:1s863y18]2[/sup:1s863y18]) = 50 ? a = 5
 
Sabzii, look up "financial formulas" ; should help you out quite a bit...
 
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