How do you write a rule for the nth term of the geometric sequence given the two terms $a_{2} = - 153, a_{4} = - 17$ $a_2=-153, a_4=-17$ ?

Question

1575 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-19T18:16:34

$u_{n} = \pm 459 \cdot {(\pm \frac{1}{3})}^{n - 1}$ $u_n = +-459 * (+-1/3)^(n-1)$

geometric sequence: $u_{n} = a r^{- 1}$ $u_n = ar^-1$

where $a$ $a$ is the starting term

and $r$ $r$ is the number by which one number is multiplied to make the next number in the sequence. (common ratio)

$u_{2} = - 153$ $u_2 = -153$
$u_{4} = - 17$ $u_4 = -17$

$u_{2} : n = 2$ $u_2: n = 2$

$u_{2} = a r^{2 - 1} = a r^{1}$ $u_2 = ar^(2-1) = ar^1$

$u_{2} = a r$ $u_2 = ar$

$u_{4} : n = 4$ $u_4: n = 4$

$u_{4} = a r^{4 - 1} = a r^{3}$ $u_4 = ar^(4-1) = ar^3$

$a r = - 153$ $ar = -153$
$a r^{3} = - 17$ $ar^3 = -17$

$r^{2} = \frac{a r^{3}}{a r} = \frac{- 17}{- 153}$ $r^2 = (ar^3)/(ar) = (-17)/-153$

$r^{2} = \frac{1}{9}$ $r^2 = 1/9$

$r = \pm \frac{1}{3}$ $r = +-1/3$

$a r = - 153$ $ar = -153$

$r = \pm \frac{1}{3}$ $r = +-1/3$

$a = - \frac{153}{\pm \frac{1}{3}} = - 153 \cdot \pm 3$ $a = -153 / ( +-1/3) = -153 * +-3$

$a = \pm 459$ $a = +-459$

the $n$ $n$ th term of the geometric sequence is $u_{n} = \pm 459 \cdot {(\pm \frac{1}{3})}^{n - 1}$ $u_n = +-459 * (+-1/3)^(n-1)$

How do you write a rule for the nth term of the geometric sequence given the two terms a_2=-153, a_4=-17a2=−153,a4=−17a_2=-153, a_4=-17?