A Quick and Easy Geometric Sequence

[@Naoseinaosei213]

I was trying to solve a similar problem,but i could not find a tutorial on how to solve it. You are great. The only thing i did not understood was the factorization.

[@ahmadmazbouh]

The hole concept of putting geometric series into equations is new for me, i'm learning from you many things 😶

[@SyberMath]

Glad to hear that!

[@chessematics]

Absolutely yes

[@tambuwalmathsclass]

My current topic, great one 👍

[@SyberMath]

Glad you liked it

[@dolevgo8535]

5:25 i know that its only one of the solutions, but i thought its pretty neat if you just compare numerators and denominators, you get 1 + r^2 + r = 7 1 + r^2 - r = 3 subtract the equations- 2r=4 => r = 2

[@advaitpetiwale9596]

But you lose the reverse progression because of r= ½.

[@jimschneider799]

Nice. Much more straightforward than my factoring (r^6 - 1)/(r^2 - 1). The only thing I would add is if you were to specify that the sequence is convergent, you could eliminate all but one possible solution.

[@brainupliftorg]

Enjoying and learning new things

[@SyberMath]

Wonderful!

[@musicsubicandcebu1774]

a1.a3 = (a2)^2 . . . substitute a1.a3 for (a2)^2 in 2nd equation. Expand (a1+a3)^2 to give (a1)^2 +2a1.a3 +(a3)^2, subtract a1.a3 to preserve the equality . . . keep substituting . . . you get the general idea.

[@PunmasterSTP]

This was a cool problem, and factoring and dividing both equations was definitely not my first guess on how to go about it. Symmetric? More like slick-metric!

[@SyberMath]

😁

[@gandharvagrover8396]

Thanks for patiently explaining beautiful answers esp the end symmetry part of two r. Keep it up :) ....

[@SyberMath]

Thank you!

[@armacham]

I was able to solve this using the quadratic formula so k^2 + k + 1 - 21/s = 0 and then do the same thing with the other equation so k^4 + k^2 + 1 - 189/s^2 which is obviously a quadratic in terms of k^2. You can apply the quadratic formula to both. from there it's just a whole lot of squaring and factoring. basically you can solve both to get k^2 + 0.5, then set them equal to one another to eliminate the k term, so you can solve for s and you find that s must be either 3 or 12 then you start over at the very beginning, but you only have two cases. Case 1, where s = 3, and case 2, where s=12. From there, it's relatively easy to solve for the value of k

[@Ni999]

The 9 got on my nerves so I took it out by setting a3=3. Three guesses later I had a solution of 12, 6, and 3. I'll come back and see the right way when I'm rested, many thanks in advance for the problem! 👍

[@SyberMath]

Absolutely!

[@quantumobject3815]

I used a bit different method Consider the sequence is a/r ,a, ar Now we substitute that in our equations we get from the second equation that a^2/r^2 + a^2+a^2.r^2=189 Now notice that we can complete the square getting that (ar+a/r)^2_a^2 = 189 From the first equation we have ar+a/r =21_a now we can simply substitute that in the other equation to get (a-21)^2 -a^2=189) No solve for a easily

[@thexavier666]

While cancellation, we should state that we are assuming it's not equal to zero. We should show that for both values of r, the expr (1+r+r^2) is not zero.

[@SyberMath]

1+r+r^2 cannot be zero for real r

[@rssl5500]

Nice syber I Alamo’s solved it love these videos :D

[@SyberMath]

Alamo’s?

[@10names55]

What is congruency/symmey of equations

[@teambellavsteamalice]

At 5:08 you can subtract giving r = (21-9/2a1 = 6/a1, isn't that faster?

[@snejpu2508]

I didn't see the cancellation. I substituted for a1, got the quartic equation. Luckily it was symetric, so I could use the famous trick for symetric quartics and you know what... I proved that there are no solutions... : ( I must have made some calculation mistake on relatively big numbers. Nice solution, by the way, but theoretically you should have solved r^2+r+1=0 to see, if there are some exclusions from the domain of the function. I this case, there are not, because it's always greater thatn 0.

[@jimmykitty]

Such a fantastic experience ❤❤🌿🌿 That was quite easy I guess 😊😸

[@mustafizrahman2822]

Einstein

[@jimmykitty]

@@mustafizrahman2822 Newton

[@mustafizrahman2822]

@@jimmykitty 😤😪😫🤨

[@advaykumar9726]

I also thought it was easy

[@SyberMath]

Yes it was!

[@krislegends]

I didn't know it was possible to divide system of equations. I wish you would have slowed down and thoroughly explained the factoring for differences of 2 squares portion.

[@SyberMath]

I will try to slow down next time. Thanks for the feedback!

[@scottleung9587]

I finally solved one before watching the video!

[@SyberMath]

Nice!

[@holyshit922]

I solved quartic Quartic which i got was easy to solve because cubic resolvent was partially factored (it was easy to pull out common factor in this cubic resolvent)

[@michaelempeigne3519]

faster way to get r value : ( 1 + r + r^2 ) - ( 1 - r + r^2 ) = 2r and if you simplify 9 / 21 = 3 / 7 so 7 - 3 = 4 so 2r = 4 == > r = 2

[@אליסושנסקי]

A very nice video as usual. I actually solved it in a different way: I squared the first equation: (a1+a2+a3)^2=a1^2+a2^2+a3^2+2(a1a2+a1a3+a2a3) If we substitute a1+a2+a3=21 and a1^2+a2^2+a3=189 and make a bit of work we get: 21^2=189+2(a1a2+a1a3+a2a3) 252=2(a1a2+a1a3+a2a3) 126=a1a2+a1a3+a2a3 126=a1*a2(1+r+r^2) As Syber Showed in his video a1*((1+r+r^2)=21 if we substitute this expression we will get that 126=21*a2, or in other words a2=6. WE know that a1+a2+a3=21 and a2=6. Subtract the second equation from the first one, and we get a1+a3=15 a3=15-a1 In Geometric series it is true that a2=sqrt(a1*a3) 6=sqrt(a1*(15-a1) Raise both sides to the second power: 36=a1^2+15*a1 a1^2-15*a1+36=0 (a1-12)(a1-3)=0 Case 1: a1=3. Than r=a2/a1=6/3=2 Case 2: a1=12. Than r=a2/a1=6/12=0.5 Job Done!!!

[@bsmith6276]

I took a different approach, solving directly for a1, a2, and a3. Because the values are in a geometric sequence then a2 is the geometric mean of a1 and a3, specifically I can write a2^2 = a1*a3. Now lets square the first given equation: (a1 + a2 + a3)^2 = 21^2. Then a bit of simplification and arrangement will give us a1^2 + a2^2 + a3^2 + 2*(a1*a2 + a1*a3 + a2*a3) = 441. Now substitute a2^2 = a1*a3 and a1^2 + a2^2 + a3^2 = 189. Then 189 + 2*(a1*a2 + a2^2 + a2*a3) = 441. We can simplify this a bit to make a2*(a1 + a2 + a3) = 126. And then another substitution gives us a2*21 = 126. Then a2=6. With a2's value known to be 6 we can go back to our original equations and substitute: a1 + 6 + a3 = 21 and a1^2 + 6^2 + a3^2 = 189 and a1*a3 = 6^2. The first and third equations simplify to a1 + a3 = 15 and a1*a3 = 36. With these we can conclude that a1 and a3 are the roots of the quadratic a^2 - 15a + 36 = 0. The quadratic a^2 - 15a + 36 = 0 is easy to solve to determine that a1 and a3 are 3 and 12 in some order. So our geometric progression is 3, 6, 12 or 12, 6, 3.

[@dennisdesormier6886]

I didn't use the geometric mean, but the rest of my solution is the same as yours. I do think that @SyberMath 's approach of adding and subtracting r^2 to do some factoring and canceling out makes his solution just a little more enjoyable, though.

[@Qermaq]

There is another Diophantine solution to the system of equations if you ignore the geometric series restriction. Thumbs up to the first comment below with that solution!

[@南投縣榮譽警察]

How can you come up with the relationship between a1 a2 a3 ? If we do not have this relationship, I think this problem would be not that easy to solve. BTW, the problems you offered are pretty interesting. Keep going on.

[@RedRad1990]

What do you mean? Like if it is not known that it's supposed to be a geometric sequence? If you know it's a geometric (or arithmetic) sequence from the beginning, it's super helpful because you immediately have a general equation (either a recurrence or a straightforward equation) for the sequence

[@MushookieMan]

It is from the definition of a geometric sequence. That was a 'given'.

[@magick333]

a simple system of 2 equations with 2 variables.

[@SyberMath]

Three variables! 😁😜

[@magick333]

@@SyberMath I haven't watched the video yet but I can solve it with just a1 and q.

[@राजनगोंगल]

👍👍👍

[@juniorochoazavalza228]

Ohh I got 11, 8, 2 still works

[@juniorochoazavalza228]

Also I got to that result in just 1 minute since I imagined that to get 9 at the end of 89 must be a result of the last number of three numbers squared. So either 5+2+2, but 2 is not a the last digit resulting of squaring any number, neither 5+3+1 since that is a problem too. So it can be the case 4+4+1. I know 64 from 8 squared so chose it to be 8. Then 144 is 12 squared but 64+144 is greater than 189 and we cannot add a negative to make it 189 since all squared real numbers are positive. So how about 4 which is just the result of 2 squared. So we now have 8 and 2 as our a1,a2. 8+2=10+a3=21 So a3=11 and if we plug in we obtain (8)^2 + (2)^2 + (11)^2 = 64 + 4 + 121 = 189

[@abhinavdiddigam2330]

You sound sick Syber, is everything ok

[@SyberMath]

Oh thanks for asking! I'm fine, possibly a minor sore throat, nothing to worry about! 💖

[@cadaver123]

🇧🇩❤🌿