In a word problem on geometric sequence where Week 1 is $50 and each week is 1.5 times the previous week, what is the explicit formula, the Week 8 amount, and the total collected over the first 8 weeks?

The sequence is geometric with $a_1=50$ and $r=1.5$, so $a_n=50(1.5)^{n-1}$, Week 8 is $50(1.5)^7\approx 854.30$, and the first-8-weeks total is $50\frac{(1.5)^8-1}{1.5-1}\approx 2462.89$.

Word Problems on Geometric Sequence: Find the n-th Term and Total

Accepted answer Answer included

Problem

A fundraiser reports that the amount collected grows by the same factor each week. Week 1 collects $50, and each subsequent week collects 1.5 times the previous week. This is a typical setup for word problems on geometric sequence.

Determine: (1) an explicit formula for the weekly amount $a_n$, (2) the Week 8 amount $a_8$, and (3) the total collected over the first 8 weeks. Monetary values are rounded to the nearest cent at the end.

Step 1: Identify the geometric sequence parameters

The phrase “each week is 1.5 times the previous week” indicates a constant common ratio. A sequence $\{a_n\}$ is geometric when $\frac{a_{n+1}}{a_n}=r$ is constant.

First term: $a_1 = 50$
Common ratio: $r = 1.5$

Recognition rule

In word problems on geometric sequence, phrases like “multiplies by,” “grows by a factor of,” “increases by a constant percent,” or “each term is a constant multiple of the previous” indicate a geometric sequence. A constant percent increase $p\%$ corresponds to $r = 1 + \frac{p}{100}$.

Step 2: Write the explicit (n-th term) formula

For a geometric sequence with first term $a_1$ and ratio $r$, the n-th term is:

\[ a_n = a_1 \cdot r^{\,n-1}. \]

Substituting $a_1=50$ and $r=1.5$ gives:

\[ a_n = 50 \cdot (1.5)^{\,n-1}. \]

Step 3: Compute the Week 8 amount

Week 8 corresponds to $n=8$:

\[ a_8 = 50 \cdot (1.5)^{7}. \]

Compute $(1.5)^7$ stepwise:

\[ (1.5)^2 = 2.25,\quad (1.5)^4 = (2.25)^2 = 5.0625,\quad (1.5)^7 = (1.5)^4 \cdot (1.5)^3 = 5.0625 \cdot 3.375 = 17.0859375. \]

Then:

\[ a_8 = 50 \cdot 17.0859375 = 854.296875 \approx 854.30. \]

Step 4: Total collected over the first 8 weeks

The total is the geometric series sum: \[ S_8 = a_1 + a_2 + \cdots + a_8. \] For $r \ne 1$, the sum of the first $n$ terms is: \[ S_n = a_1 \cdot \frac{r^n - 1}{r - 1}. \]

With $a_1=50$, $r=1.5$, and $n=8$:

\[ S_8 = 50 \cdot \frac{(1.5)^8 - 1}{1.5 - 1}. \]

Since $(1.5)^8 = (1.5)^7 \cdot 1.5 = 17.0859375 \cdot 1.5 = 25.62890625$,

\[ S_8 = 50 \cdot \frac{25.62890625 - 1}{0.5} = 50 \cdot \frac{24.62890625}{0.5} = 50 \cdot 49.2578125 = 2462.890625 \approx 2462.89. \]

Summary of results

Quantity	Exact expression	Approximate value
Explicit formula	$a_n = 50 \cdot (1.5)^{n-1}$	—
Week 8 amount	$a_8 = 50 \cdot (1.5)^7$	$\approx 854.30$
Total of Weeks 1–8	$S_8 = 50 \cdot \frac{(1.5)^8 - 1}{1.5 - 1}$	$\approx 2462.89$

Visualization

The plotted points show the weekly amounts $a_n = 50 \cdot (1.5)^{n-1}$. The curve rises because the common ratio $r=1.5>1$, a hallmark of geometric growth in real-world word problems on geometric sequence.

Final answers

$a_n = 50 \cdot (1.5)^{n-1}$
$a_8 \approx 854.30$
$S_8 \approx 2462.89$

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Quantity	Exact expression	Approximate value
Explicit formula	\(a_n = 50 \cdot (1.5)^{n-1}\)	—
Week 8 amount	\(a_8 = 50 \cdot (1.5)^7\)	\(\approx 854.30\)
Total of Weeks 1–8	\(S_8 = 50 \cdot \frac{(1.5)^8 - 1}{1.5 - 1}\)	\(\approx 2462.89\)