SOLUTION
We were given the geometric sequence 16, 24, 36, 54 and we are told to find
[tex]S_{12}[/tex]
This means to find the sum of the first 12 terms.
From the sequence
[tex]\begin{gathered} \text{the first term a = 16} \\ \text{the common ratio r = }\frac{24}{16}=\frac{3}{2} \\ n\text{umber of terms n = 12} \end{gathered}[/tex]
Sum of a geometric sequence is given as
[tex]\begin{gathered} S_n=\frac{a(r^n-1)}{r-1} \\ \text{since r }>\text{ 1} \end{gathered}[/tex]
So, we have
[tex]\begin{gathered} S_n=\frac{a(r^n-1)}{r-1} \\ S_{12}=\frac{16((\frac{3}{2})^{12}-1)}{\frac{3}{2}-1} \\ S_{12}=\frac{16(128.7464)}{0.5} \\ S_{12}=4119.8848 \end{gathered}[/tex]
Hence to the nearest whole number, the answer is 4,120 second option
