Question

What is the sum of the geometric series in which a1 = 4, r = 3, and an = 324?

Hint: cap s sub n equals start fraction a sub one left parenthesis one minus r to the power of n end power right parenthesis over one minus r end fraction comma r ≠ 1, where a1 is the first term and r is the common ratio.

Answer 1

The sum of the geometric series is approximately 324.

Step-by-step explanation:

The sum of a geometric series can be found using the formula:

Sn = a1(1 - rn) / (1 - r)

Where Sn is the sum of the series, a1 is the first term, r is the common ratio, and n is the number of terms.

Plugging in the given values, we have:

Sn = 4(1 - 3n) / (1 - 3)

Now we can solve for n by plugging in the given value of an:

324 = 4(1 - 3n) / (1 - 3)

Cross-multiplying and simplifying:

324(1 - 3) = 4(1 - 3n)

324 - 972 = 4 - 12n

-648 = -8n

Dividing both sides by -8, we get:

81 = 2n

Taking the logarithm base 2 of both sides:

n = log2(81)

Using a calculator, we find that n is approximately 6.3398.

Now we can plug this value back into the formula:

Sn = 4(1 - 36.3398) / (1 - 3)

Simplifying further:

Sn ≈ -648 / -2

Sn ≈ 324

Answer 2

`\bf n^(th)\textit{ term of a geometric sequence}\\\\ a_n=a_1\cdot r^(n-1)\qquad \begin{cases} n=n^(th)\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ a_1=4\\ r=3\\ a_n=324 \end{cases} \implies 324=4(3)^(n-1) \\\\\\ \cfrac{324}{4}=3^(n-1)\implies 81=3^(n-1)\implies 3^4=3^(n-1)\implies 4=n-1 \\\\\\ \boxed{5=n}\\\\`


`\bf -------------------------------\\\\ \qquad \qquad \textit{sum of a finite geometric sequence}\\\\ S_n=\sum\limits_(i=1)^(n)\ a_1\cdot r^(i-1)\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=n^(th)\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ a_1=4\\ r=3\\ n=5 \end{cases} \\\\\\ S_5=4\left( \cfrac{1-3^5}{1-3} \right)\implies S_5=4\left(\cfrac{1-243}{-2} \right)`

and surely you know how much that is.