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