Easy G.P question=help? (1 Viewer)

[LoveHateSchool]

The sum if the first 5 terms of a G.P is 77 and the sum of the next 5 tearms is -2464. Find the 4th term of the series.

REP and virtual brownie points for help.

 

[barbernator]

i see you use margaret grove

 

[Carrotsticks]

Do you want us to do the whole question for you, or point you in the right direction?

 

[LoveHateSchool]

^Is that a textbook? It was just off my holiday revision sheets lol

 

[Carrotsticks]

^Is that a textbook? It was just off my holiday revision sheets lol

Margaret Grove is the author of the commonly used Maths in Focus textbooks.

 

[LoveHateSchool]

^^Well I thought it had to with how you can write T1 as a and that up to 5, and then the other one is T6-T10 and then divide them out somehow, use them simultaneously but I really don't know.

Maybe if I saw the start of the working the rest would click?

OH and never seen a MIF textbook, we use Couchman and Jones From like the 1980s

 

[barbernator]

yeh, MIF textbook is where that question comes from. I'll let carrot do the question for you anyway

 

[deswa1]

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{a(r^5-1)}{r-1}=77\\\frac{a(r^{10}-1)}{r-1}=-2387\\ \frac{r^{10}-1}{r^5-1}=-31\\ \frac{(r^5@plus;1)(r^5-1)}{r^5-1}=-31\\ r^5@plus;1=-31\\ r=-2\\ \frac{a(-32-1)}{-2-1}=77 ->a=7\\ T_{4}=ar^3=7(-2)^3=-56" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{a(r^5-1)}{r-1}=77\\\frac{a(r^{10}-1)}{r-1}=-2387\\ \frac{r^{10}-1}{r^5-1}=-31\\ \frac{(r^5+1)(r^5-1)}{r^5-1}=-31\\ r^5+1=-31\\ r=-2\\ \frac{a(-32-1)}{-2-1}=77 ->a=7\\ T_{4}=ar^3=7(-2)^3=-56" title="\frac{a(r^5-1)}{r-1}=77\\\frac{a(r^{10}-1)}{r-1}=-2387\\ \frac{r^{10}-1}{r^5-1}=-31\\ \frac{(r^5+1)(r^5-1)}{r^5-1}=-31\\ r^5+1=-31\\ r=-2\\ \frac{a(-32-1)}{-2-1}=77 ->a=7\\ T_{4}=ar^3=7(-2)^3=-56" /></a>

EDIT: A bit of explanation. The RHS in the second line comes from adding the first 5 terms and the next 5 terms, to give the first 10 terms, otherwise the formula doesn't work. The third line comes from dividing the first two lines in order to eliminate the a's and allow us to solve for r.

 

[Carrotsticks]

yeh, MIF textbook is where that question comes from. I'll let carrot do the question for you anyway

Looks like deswa1 beat me to it =p

A tip:

You can use the following formula for finding the sum, but starting from the m'th term instead of the 0'th (or 1'st) term.



You can derive it yourself by acquiring the formulas for S_n and S_m, then subtracting S_m from S_n.

 

[deswa1]

Whoops, sorry Carrot... Would you recommend doing it using that formula or the way I did it by combining the first 10 terms together and then staring from the first term?

 

[Carrotsticks]

Whoops, sorry Carrot... Would you recommend doing it using that formula or the way I did it by combining the first 10 terms together and then staring from the first term?

Of course using the intuitive method! You know how much I hate rote learning.

But perhaps adding a bit of explanation to your solution (ie: you acquired your third line by dividing the first two expressions) would make things a lot easier for your marker.

 

[LoveHateSchool]

Thank you guys, I have to wait before I can rep you deswa.

I understand the way deswa did it, I don't understand the formula, would the mth term in this Q be the 10?
It's probably jsut easier to manipulate the formulas I know to find unknowns heh

 

[Carrotsticks]

Hold on, I actually interpreted the question differently.

I interpreted it as:

"The first 5 terms has sum 77, and the NEXT 5 terms (so not including the first 5) has sum -2464"

So:



Hence why I introduced that formula.

But since the numbers worked out so nicely in deswa1's solution, I'm assuming that was Margaret's intention.

 

[deswa1]

Hold on, I actually interpreted the question differently.

I interpreted it as:

"The first 5 terms has sum 77, and the NEXT 5 terms (so not including the first 5) has sum -2464"

So:



Hence why I introduced that formula.

But since the numbers worked out so nicely in deswa1's solution, I'm assuming that was Margaret's intention.

Yeah, your interpretation is right. What I did though was since the first five terms sum to 77 and the next 5 sum to -2464, the first 10 sum to -2387 by adding them together.

 

[Carrotsticks]

Haha, I did not notice that the number was -2387 as opposed to -2464.

I am going blind.

 

[RivalryofTroll]

Of course using the intuitive method! You know how much I hate rote learning.

But perhaps adding a bit of explanation to your solution (ie: you acquired your third line by dividing the first two expressions) would make things a lot easier for your marker.

Haters gonna hate! (jokes)

Yeah, I like Deswa's method

Question about G.P.
Sn = a(1-r^n)/1-r is also valid right?

 

[deswa1]

Haters gonna hate! (jokes)

Yeah, I like Deswa's method

Question about G.P.
Sn = a(1-r^n)/1-r is also valid right?

Of course. The only thing different about both formulas is that you multiply through by -1. Since though you multiply the denominator and the numerator by the same thing, you aren't changing the value of the fraction. Some people like using one or the other depending on the size of the number (i.e. use 1-r if r is less than 1) but I always use r-1 for everything. It doesn't matter though.

 

[RivalryofTroll]

Of course. The only thing different about both formulas is that you multiply through by -1. Since though you multiply the denominator and the numerator by the same thing, you aren't changing the value of the fraction. Some people like using one or the other depending on the size of the number (i.e. use 1-r if r is less than 1) but I always use r-1 for everything. It doesn't matter though.

hmmmm...

When you changed r^10 -1 into (r^5 - 1)(r^5 +1) to cancel down, I realised I have a long way to go until I get those mathematical instincts
I understood everything else though

 

[bleakarcher]

LOL, I completely forgot about the formula for the sum of the first n terms. I did it this way.

Let the first term be a, common ratio be b.
a+ab+ab^2+ab^3+ab^4=77 (1)
ab^5+ab^6+ab^7+ab^8+ab^9=-2464 (2)
From (2): b^5[a+ab+ab^2+ab^3+ab^4]=-2464
i.e. 77b^5=-2464
b=-2
From (1): a=77/[1+b+b^2+b^3+b^4]=77/[1+(-2)+(-2)^2+(-2)^3+(-2)^4]=7
So the fourth term ab^3=7*(-2)^3=7*-8=-56

 

[deswa1]

Yeah, I was thinking about doing it like that (its essentially the same thing) but I try to minimise the number of terms I have to write out in latex because I'm pretty slow haha.