Question

Find an explicit rule for the nth term of the sequence. 3, -12, 48, -192, ...

Answer 1

Answer: $a_n=3(-4)^{n-1}$

Step-by-step explanation:

Given sequence : $3, -12, 48, -192, ...$

Here, First term $a_1=3$

Second term $a_2=-12$

and  Ratio=$r=\frac{a_2}{a_1}=\frac{-12}{3}=-4$

Third term $a_3=48$

Ratio=$r=\frac{a_3}{a_2}=\frac{48}{-12}=-4$

Fourth term $a_4=-192$

Ratio=$r=\frac{a_4}{a_3}=\frac{-192}{48}=-4$

Thus, the given sequence is geometric sequence with the common ratio r = -4.

The explicit rule for for the nth term in geometric sequence is given by

$a_n=a_{1}r^{n-1}$

Then the explicit rule for the nth term of the given sequence will be :-

$a_n=3(-4)^{n-1}$

Answer 2

Hello,
$u_{1}=3$
$u_{2}=3*(-4)$
$u_{3}=3*(-4)^2$

$u_{n}=3*(-4)^{n-1}$