Find the tenth term of the sequence a^2/b, -a, b, -b^2/a

Question

This is for the geometric sequences hmwk and I'm just confused on what is even happening I cannot see what Thier multiplying it by for the eqn t of n=a×r^n-1  I know a is a^2 over b, but what is r(the common ratio) for this sequence?

Answer 1

Note that there are two terms that make it easy to determine the common ratio r:   (b)/(-a).  But, be sure to check that with the other terms.

 

So, the first term is a²/b.  The common ratio is (-b/a).  What is the tenth term?

 

    T_n = T₁*r^(n-1)                 where T₁=a²/b   and   r=(-b/a)

 

    T₁₀ = T₁*r^(10-1)

    T₁₀ = (a²/b)*(-b/a)⁹

    T₁₀ = -b⁸/a⁷

 

You could check the answer by continuing to list terms in the given sequence.

Answer 2

Just for the ratios without thinking too hard:

Second/First ratio  =  -a / (a²/b)

  (convert the first "/" into a multiply by multiplying by the reciprocal)

    = -a * (b/a²) = -ab/a² = -b/a

 

Third/Second ratio = b / (-a) = -b/a   same ratio

 

Fourth/Third = (-b²/a) / b = -b2/a* 1/b = -b2/(ab) = -b/a

 

So the common ratio is (-b/a)