Question

in a geometric sequence, $a_4 = 54$ and $a_7 = 1,458$. what is the 12th term?\
\bigcirc\\ 5,616\
\bigcirc\\ 354,294\
\bigcirc\\ 14,350,365\
\bigcirc\\ 234,812,358

Explanation:

Step1: Recall the formula for a geometric sequence

The nth term of a geometric sequence is given by \( a_n = a_1 r^{n - 1} \), where \( a_1 \) is the first term and \( r \) is the common ratio. We know that \( a_4 = 54 \) and \( a_7 = 1458 \). So we can write:
\( a_4 = a_1 r^{4 - 1}=a_1 r^3 = 54 \) (Equation 1)
\( a_7 = a_1 r^{7 - 1}=a_1 r^6 = 1458 \) (Equation 2)

Step2: Find the common ratio \( r \)

Divide Equation 2 by Equation 1:
\( \frac{a_1 r^6}{a_1 r^3}=\frac{1458}{54} \)
Simplify the left side: \( r^{6 - 3}=r^3 \)
Simplify the right side: \( \frac{1458}{54} = 27 \)
So \( r^3 = 27 \), then \( r=\sqrt[3]{27}=3 \)

Step3: Find the first term \( a_1 \)

Substitute \( r = 3 \) into Equation 1: \( a_1(3)^3 = 54 \)
\( 27a_1 = 54 \)
Solve for \( a_1 \): \( a_1=\frac{54}{27}=2 \)

Step4: Find the 12th term \( a_{12} \)

Use the formula \( a_n = a_1 r^{n - 1} \) with \( n = 12 \), \( a_1 = 2 \), and \( r = 3 \):
\( a_{12}=2\times3^{12 - 1}=2\times3^{11} \)
Calculate \( 3^{11}=177147 \)
Then \( a_{12}=2\times177147 = 354294 \)

Answer:

354,294