Question

question
find the sum of the first 7 terms of the following sequence. round to the nearest hundredth if necessary.
32, 40, 50, ...
sum of a finite geometric series:
$s_n = \frac{a_1 (1 - r^n)}{1 - r}$

Explanation:

Step1: Identify the common ratio \( r \)

To find the common ratio \( r \) of a geometric sequence, we divide a term by its previous term.
For the given sequence \( 32, 40, 50, \dots \),
\( r=\frac{40}{32} = 1.25\) and also \( r=\frac{50}{40}=1.25 \). So the common ratio \( r = 1.25\), the first term \( a_1=32 \) and the number of terms \( n = 7\).

Step2: Apply the finite geometric series formula

The formula for the sum of the first \( n \) terms of a geometric series is \( S_{n}=\frac{a_{1}(1 - r^{n})}{1 - r} \) (when \( r
eq1 \)).
Substitute \( a_1 = 32 \), \( r=1.25 \), and \( n = 7 \) into the formula:
First, calculate \( r^{n}=(1.25)^{7}\).
\( (1.25)^{7}=1.25\times1.25\times1.25\times1.25\times1.25\times1.25\times1.25 \)
\( 1.25\times1.25 = 1.5625 \)
\( 1.5625\times1.25=1.953125 \)
\( 1.953125\times1.25 = 2.44140625 \)
\( 2.44140625\times1.25=3.0517578125 \)
\( 3.0517578125\times1.25 = 3.814697265625 \)
\( 3.814697265625\times1.25=4.76837158203125 \)
Then, \( 1-r^{n}=1 - 4.76837158203125=- 3.76837158203125 \)
\( 1 - r=1 - 1.25=-0.25 \)
Now, \( S_{7}=\frac{32\times(1-(1.25)^{7})}{1 - 1.25}=\frac{32\times(- 3.76837158203125)}{-0.25} \)
First, calculate the numerator: \( 32\times(-3.76837158203125)=-120.587890625 \)
Then, divide by the denominator: \( \frac{- 120.587890625}{-0.25}=482.3515625 \)

Answer:

\( 482.35 \) (rounded to the nearest hundredth)