Study buddies! (1 Viewer)

[tommykins]

You may do so on boys, not chicks as they are already reserved.

We'll fight to the death of solve numerous mathematical questions over it.

 

[foram]

Okay solve this:
(i'll post the answer later.)
Three numbers are consecutive terms in a geometric series.
Their sum is 58.5 and their product is 3375.

i) Write down a relationship (using the symbols a and r) which shows that the numbers are consecutive terms in a geometric series.

ii) Find the three numbers.

 

[lyounamu]

We'll fight to the death of solve numerous mathematical questions over it.

No, that's unfair!!! I will get nailed by you!!!

Why not do something that will test our manliness???

 

[foram]

why not do chemistry to the death?

 

[tommykins]

No, that's unfair!!! I will get nailed by you!!!

Why not do something that will test our manliness???

Both our penis sizes will require microscopes so that's out of it.

Fight to the death.

 

[lyounamu]

Okay solve this:
(i'll post the answer later.)
Three numbers are consecutive terms in a geometric series.
Their sum is 58.5 and their product is 3375.

i) Write down a relationship (using the symbols a and r) which shows that the numbers are consecutive terms in a geometric series.

ii) Find the three numbers.

1) a, ar and ar^2

2) r = 2.5 and a = 6
6, 15 and 37.5


I cannot be bothered to put up my 20 lines of working but I used simultaneous equation where I went a^3 . r^3 = 3375
(a(1 + r +r^2))^3 = 58.5^3

 

[lyounamu]

Both our penis sizes will require microscopes.

I am not sure about that.

 

[tommykins]

I am not sure about that.

Yeah, true - it'd just be yours.

Three numbers are consecutive terms in a geometric series.
Their sum is 58.5 and their product is 3375.

i) Write down a relationship (using the symbols a and r) which shows that the numbers are consecutive terms in a geometric series.

ii) Find the three numbers.

i ) a, ar, ar²
ii) a³r³ = 3375 => ar = 15 let r = 5/a

a+ar+ar² = 58.5

a+a(5/a) + a(5/a)² = 58.5

a + 5 + a(25/a²) = 58.5
a+5+25/a = 58.5

????

Answers here.

 

[x.Exhaust.x]

We'll fight to the death of solve numerous mathematical questions over it.

Lol. I'll fight you with a shuffle battle [I've seen your sig ).

 

[lyounamu]

1) a, ar and ar^2

2) r = 2.5 and a = 6
6, 15 and 37.5


I cannot be bothered to put up my 20 lines of working but I used simultaneous equation where I went a^3 . r^3 = 3375
(a(1 + r +r^2))^3 = 58.5^3

If you want me to post up working out I will be glad to but it is a quite a long working out.

 

[tommykins]

Lol. I'll fight you with a shuffle battle [I've seen your sig ).

Oh god, shuffle battles are for like the biggest losers out there.

I'd rip you anyways, PM me if you really want to establish my awesomeness.

 

[lyounamu]

Did I get it right or what? I think I did. If I did, I will post my working out since it will be pointless to put one if I got it wrong.

 

[tommykins]

Did I get it right or what? I think I did. If I did, I will post my working out since it will be pointless to put one if I got it wrong.

Your solutions are correct.

My approach was a³r³ = 3375 => ar = 15
a = 15/r

sub it into a+ar+ar², solve quadratic, resub into ar = 15.

 

[lyounamu]

Your solutions are correct.

My approach was a³r³ = 3375 => ar = 15
a = 15/r

sub it into a+ar+ar², solve quadratic, resub into ar = 15.

Thanks. Here is my working out if anyone didn't know

Let the 3 numbers be a, ar and ar^2

The product of 3 numbers is: a^3 . r^3 = 3375 ....(1)
The sum of 3 numbers is: a + ar + ar^2 = a(1+r+r^2) = 58.5 ... (2)

(2)^3: a^3(1+r+r^2)^3 = 58.5^3 (200202.625) ... (3)
(3) divided by (2): (1+r+r^2)^3 / r^3 = 59.319
1 + r +r^2 / r = 3.9
r^2 +1+r = 3.9r
r^2 -2.9r +1 = 0
Therefore, r = 2.5 (I used the quadratic formula and r > 0)

I substituted r = 2.5 into the equation (1) to find what a is equal to.

It was an easy question. I thank you for great revision question, Foram!!!

 

[tommykins]

Thanks. Here is my working out if anyone didn't know

Let the 3 numbers be a, ar and ar^2

The product of 3 numbers is: a^3 . r^3 = 3375 ....(1)
The sum of 3 numbers is: a + ar + ar^2 = a(1+r+r^2) = 58.5 ... (2)

(2)^3: a^3(1+r+r^2)^3 = 58.5^3 (200202.625) ... (3)
(3) divided by (2): (1+r+r^2)^3 / r^3 = 59.319
1 + r +r^2 / r = 3.9
r^2 +1+r = 3.9r
r^2 -2.9r +1 = 0
Therefore, r = 2.5 (I used the quadratic formula and r > 0)

I substituted r = 2.5 into the equation (1) to find what a is equal to.

It was an easy question. I thank you for great revision question, Foram!!!

Of course you got the answer, but watch out with those decimals, try to do the most simplest working out as to prevent any unncessary errors.

 

[lyounamu]

Of course you got the answer, but watch out with those decimals, try to do the most simplest working out as to prevent any unncessary errors.

Thanks. I actually did mistake during my working out (with the decimal ones!!!, what a prediction!). I better do another 100s of those over again not to make a mistake.

 

[foram]

Solution:

Let the terms be a/r , a , and ar

Therefore the product is a^3 = 3375
Therefore a=15

Sum= a/r + r + ar = 58.5

=15/r + 15 + 15r = 58.5

15 + 15r + 15r^2 = (58.5)r

15r^2 -43.5r +15 = 0

r= 2.5 or 0.5

Hence series is 6, 15, 37.5 or 37.5, 15, 6

If the series is written as a, ar, ar^2

then (ar)^3

ar=15
a=15/r

15/r + 15 + 15r = 58.5 as stated before.

EDIT: Good work solving it. It was on my 2U half yearly. It was a few easy marks for me, but other people couldn't do it. When I did it, I let the series be a/r, a, ar, but I guess it works the other way too.

 

[tania O]

i know ayyy lol i live in fairfield lol

 

[lyounamu]

Solution:

Let the terms be a/r , a , and ar

Therefore the product is a^3 = 3375
Therefore a=15

Sum= a/r + r + ar = 58.5

=15/r + 15 + 15r = 58.5

15 + 15r + 15r^2 = (58.5)r

15r^2 -43.5r +15 = 0

r= 2.5 or 0.5

Hence series is 6, 15, 37.5 or 37.5, 15, 6

If the series is written as a, ar, ar^2

then (ar)^3

ar=15
a=15/r

15/r + 15 + 15r = 58.5 as stated before.

EDIT: Good work solving it. It was on my 2U half yearly. It was a few easy marks for me, but other people couldn't do it. When I did it, I let the series be a/r, a, ar, but I guess it works the other way too.

That's pretty smart too!

By the way, how did you go in that task???

 

[yeeshu]

I reckon this thread is dead. It is tragic that this thread passed away...

This thread may pass away...but sum things never give way...and go on...forever and ever