Answer :

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

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