Question

one term of a geometric sequence is $a_7 = 1458$. the common ratio is $r = 3$. write the first term for the sequence.
a $a_1 = 3$
b $a_1 = 2$
c $a_1 = \frac{1}{2}$
d $a_1 = 4$

Explanation:

Step1: Recall the formula for a geometric sequence

The formula for the \(n\)-th term of a geometric sequence is \(a_n = a_1 \cdot r^{n - 1}\), where \(a_n\) is the \(n\)-th term, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.

Step2: Substitute the given values into the formula

We know that \(a_7 = 1458\), \(r = 3\), and \(n = 7\). Substituting these into the formula \(a_n = a_1 \cdot r^{n - 1}\), we get:
\[
1458 = a_1 \cdot 3^{7 - 1}
\]

Step3: Simplify the exponent

Simplify \(3^{7 - 1}\) to \(3^6\). We know that \(3^6 = 729\), so the equation becomes:
\[
1458 = a_1 \cdot 729
\]

Step4: Solve for \(a_1\)

To find \(a_1\), divide both sides of the equation by 729:
\[
a_1 = \frac{1458}{729}
\]
Simplifying \(\frac{1458}{729}\) gives \(a_1 = 2\).

Answer:

B. \(a_1 = 2\)