A geometric sequence is one in which consecutive terms form a fixed ratio r. In other words, if aₙ is the nth term in the sequence, then
[tex]a_n = a_{n-1}r[/tex]
For example, if a₁ = a is the 1st term, then
2nd term = a₂ = a₁r
3rd term = a₃ = a₂r = a₁r²
4th term = a₄ = a₃r = a₁r³
and so on. It's fairly easy to infer that
nth term = [tex]a_n = a_{n-1}r = a_{n-2}r^2 = a_{n-3}r^3 = \cdots = a_1r^{n-1}[/tex]
14. We're given the 2nd and 5th terms, a₂ = -243 and a₅ = -9, and we use them to find the ratio r.
a₅ = a₄r = a₃r² = a₂r³
-9 = -243 r³
r³ = 1/27
⇒ r = 1/3
Then the 1st term is
a₁ = a₂/r = -243/(1/3) = -729
and the nth term is recursively given by
[tex]a_n = \dfrac13a_{n-1}[/tex]
and explicitly by
[tex]a_n = \left(\dfrac13\right)^{n-1} a_1 = -\dfrac{729}{3^{n-1}} = -\dfrac{3^6}{3^{n-1}} = -3^{7-n}[/tex]
15. Now we have a₄ = 1/72 and a₃ = -1/12. Using what we know about geometric sequences, we have
a₄ / a₃ = (a₃r) / a₃ = r
so that
r = (1/72) / (-1/12) = -1/6
Then the 1st term is
a₁ = a₂/r = a₃/r² = (-1/12) / (-1/6)² = -3
and the nth term is recursively given by
[tex]a_n = -\dfrac16a_{n-1}[/tex]
and explicitly by
[tex]a_n = \left(-\dfrac16\right)^{n-1} (-3) = -3\cdot(-6)^{1-n} = 3\cdot(-1)^n\cdot6^{1-n}[/tex]
