What is the value of m in the given geometric sequence?

  • Thread starter Thread starter hostergaard
  • Start date Start date
  • Tags Tags
    Sequence
AI Thread Summary
The discussion centers on finding the value of m in a geometric sequence formed by three terms from an arithmetic sequence defined by the sum formula sn=4n²-2n. To find the terms, the nth term can be calculated using un = sn - sn-1. Specifically, u2 and u32 are derived from the sum formula by calculating s2, s1, s32, and s31, then subtracting accordingly. The relationship between the terms in the geometric sequence is established through the equation u_m/u2 = u32/u_m. The conversation emphasizes the need to apply properties of geometric sequences to solve for m.
hostergaard
Messages
37
Reaction score
0
the sum of the first n terms of an arithmetic sequence {un} is given by the
formula sn=4n2-2n. three terms of this secuence, u2 um and u32 are conscutive terms en a geomtric sequence. find m.

Yea, I am really confused... please help :-)
 
Physics news on Phys.org
you can find tn by

u_{n} = s_{n} - s_{n-1}

Then find u_{m} by applying properties of geometric sequence. work back and find m
 
In particular, u_2= s_2- s_1. Find s_2 and s_1 from the formula you are given and subtract. u_{32}= s_{32}- s_{31}. Again find those and subtract. Now you know two terms of a geometric sequence you can find the term between them. \frac{u_m}{u_2}= \frac{u_{32}}{u_m}.
 
Last edited by a moderator:
Helpiamtrappedinatextagandican'tescapefromit!
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Back
Top