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C2 Geometric

In a GP the 2nd term is -12 and the 5th term is 768. Find the common ratio and the first term. how would you do this one?
Reply 1
By dividing the terms
Reply 2
Original post by Secreay
In a GP the 2nd term is -12 and the 5th term is 768. Find the common ratio and the first term. how would you do this one?


We know the formula is nth term=arn1 \text{nth term} = ar^{n-1}

Sub in -12 and n=2 in the formula, then sub in 768 and n=5, you will get two equations, solve them simultaneously.
Reply 3
-12/r = 768/r^4 seems wrong :/
Reply 4
Original post by Secreay
-12/r = 768/r^4 seems wrong :/


12r=768r4    r4r=76812 \displaystyle \frac{-12}r = \frac{768}{r^4} \implies \frac{r^4}r = - \frac{768}{12}

You are going right.
Reply 5
but i cant cube root i neg number
Reply 6
Original post by Secreay
but i cant cube root i neg number


why not?
Reply 7
Original post by Secreay
but i cant cube root i neg number


Negative numbers can be cube rooted but you can't take the square root of them. Input it in your calc, and it will give the answer.
Reply 8
HAHA im a noob so r=-4?
Reply 9
Original post by Secreay
HAHA im a noob so r=-4?


:yep:
Original post by raheem94
12r=768r4    r4r=76812 \displaystyle \frac{-12}r = \frac{768}{r^4} \implies \frac{r^4}r = - \frac{768}{12}

You are going right.

This is how I went around the equation.

You might find this way easier ^^

-12r^-1=768r^-4
-12=768r-4/r^-1
-12=(768/r^4)/(1/r)
-12=(768/r^4)*(r/1) [doing the dividing fractions so flip the right]
-12=768r/r^4 [now apply indices rules here]
-12=768r^-3
-1/64=r^-3
-1/64=1/r^3
-64/1=r^3/1 [flipped top and bottom]
-4=r
Reply 11
Original post by Stickyelmo
This is how I went around the equation.

You might find this way easier ^^

-12r^-1=768r^-4
-12=768r-4/r^-1
-12=(768/r^4)/(1/r)
-12=(768/r^4)*(r/1) [doing the dividing fractions so flip the right]
-12=768r/r^4 [now apply indices rules here]
-12=768r^-3
-1/64=r^-3
-1/64=1/r^3
-64/1=r^3/1 [flipped top and bottom]
-4=r


I really don't understand, why you quoted me?

Its not a problem for me to deal with such questions and the way i was doing was easier than your's.

r4r=76812    r3=64    r=643=4 \displaystyle \frac{r^4}r = - \frac{768}{12} \implies r^3 = -64 \implies r = \sqrt[3]{-64} = -4

Who's method is better?
Original post by raheem94
I really don't understand, why you quoted me?

Its not a problem for me to deal with such questions and the way i was doing was easier than your's.

r4r=76812    r3=64    r=643=4 \displaystyle \frac{r^4}r = - \frac{768}{12} \implies r^3 = -64 \implies r = \sqrt[3]{-64} = -4

Who's method is better?


his method is better....less booky :tongue:
Reply 13
Both methods are exactly the same.. one looks shorter as a lot of the steps can be done mentally and not written down.
Reply 14
Original post by David_Skiller
his method is better....less booky :tongue:


Original post by F1Addict
Both methods are exactly the same.. one looks shorter as a lot of the steps can be done mentally and not written down.


Correct, both are the same, but i was surprised to be quoted, may be if the post would have been directed to the OP then it would have made a bit of sense.
Original post by raheem94
Correct, both are the same, but i was surprised to be quoted, may be if the post would have been directed to the OP then it would have made a bit of sense.



"Who's method is better?"

you were asking the question faget :tongue:

go back to doing questions in the book :biggrin:
ah woops I quoted the wrong text haha X_X

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