Question

What is the 6th term of the geometric sequence f(1) = 12, f(n) = -2/3f(n-1)

Answer 1

$\bf \begin{array}{ccll} \stackrel{n^{th}}{term}&value\\\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\\\ 1&12\\\\ 2&-\frac{2}{3}f(2-1)\\\\ &-\frac{2}{3}f(1)\\\\ &-\frac{2}{3}(12)\\\\ &-8\\\\ 3&-\frac{2}{3}f(3-1)\\\\ &-\frac{2}{3}f(2)\\\\ &-\frac{2}{3}(-8)\\\\ &\frac{16}{3} \end{array}$

$\bf \begin{array}{cccl} \qquad &\qquad \\ 4&-\frac{2}{3}f(4-1)\\\\ &-\frac{2}{3}f(3)\\\\ &-\frac{2}{3}\left( \frac{16}{3} \right)\\\\ &-\frac{32}{9}\\\\ 5&-\frac{2}{3}f(5-1)\\\\ &-\frac{2}{3}f(4)\\\\ &-\frac{2}{3}\left( -\frac{32}{9} \right)\\\\ &\frac{64}{27}\\\\ 6&-\frac{2}{3}f(6-1)\\\\ &-\frac{2}{3}f(5)\\\\ &-\frac{2}{3}\left( \frac{64}{27}\right) \end{array}$