Verified Answer :

dalendrk

\(-3;-12;-48;-192;...\\\\It's\ the\ geometric\ sequence:a_n=a_1r^{n-1}\\\\a_1=-3;\ a_2=-12\\\\r=a_2:a_1\to r=-12:(-3)=4\\\\\boxed{a_n=-3\cdot4^{n-1}}\to\fbox{b.}\)

Answer:

b)  \(a_{n} = (-3)4^{n-1}\).

Step-by-step explanation:

Given :  -3, -12, -48, -192, ...

To find : find an equation for the nth term of the sequence.

Solution : We have given  -3, -12, -48, -192, ...

Nth term of geometric sequence = \(a_{n} = a_{1}r^{n-1}\).

Where, \(a_{n}= nth terms\).

\(a_{1}= first term\).

r = common ratio.

r = \(\frac{second term}{first term}\)

r =  \(\frac{-12}{-3}\) .

r = 4.

\(a_{n} = (-3)4^{n-1}\).

Therefore,b)  \(a_{n} = (-3)4^{n-1}\).

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