Determine the sum of the first seven terms of the geometric series in which the third term is 18 and the terms increase by a factor of 3
AAAaannndd the bot gets it wrong yet again!
Once again, botching even simple math.
ar^2=18 and r=3, so a=18/3^2 = 2
so S7 = a(r^7-1)/(r-1) = 2(3^7 - 1)/(3-1) = 2186
I apologize for the incorrect solution. You are absolutely correct that $a=2$ and $r=3$ in this problem, and using the formula for the sum of a geometric series gives:
$S_7=\frac{ar^7-a}{r-1}=\frac{2 \cdot 3^7-2}{3-1}=2186$.
Thank you for pointing out the mistake.
To determine the sum of the first seven terms of a geometric series, we'll need to find the first term (a) and the common ratio (r).
We are given that the third term is 18 and the terms increase by a factor of 3.
Let's list out the terms of the series to see the pattern:
1st term: a
2nd term: ar (since the terms increase by a factor of r)
3rd term: ar^2 = 18
From the information given, we can deduce that ar^2 = 18.
Now, we need to find the common ratio (r). Let's divide the 3rd term by the 2nd term:
(ar^2)/(ar) = r
Simplifying, we get:
r = 18/1 = 18
Now, we can find the first term (a) by dividing the 3rd term by the square of the common ratio (r):
a = (3rd term)/(r^2) = 18/(18^2) = 18/324 = 1/18
So, the first term (a) is 1/18 and the common ratio (r) is 18.
To find the sum of the first seven terms, we'll use the formula for the sum of a finite geometric series:
S = a(r^n - 1)/(r - 1)
where S is the sum, a is the first term, r is the common ratio, and n is the number of terms.
Plugging in the values, we have:
S = (1/18)((18^7 - 1)/(18 - 1))
Simplifying the value inside the brackets, we get:
S = (1/18)(669,615,201 - 1)/17)
Simplifying further, we have:
S = (1/18)(669,615,200/17)
Finally, we calculate the sum:
S ≈ (1/18)(39,386,782.35)
S ≈ 2,188,154.58
Therefore, the sum of the first seven terms of the geometric series is approximately 2,188,154.58.
Let the first term be $a$ and the common ratio be $r$. We know that the third term is 18, so $ar^2=18$. We also know that the terms increase by a factor of 3, so $ar^4=54$. Dividing these two equations gives $r^2=3$, so $r=\sqrt{3}$ (we reject the negative root because the terms are increasing). Substituting into the first equation gives $a=\frac{18}{3}=6$.
The sum of the first seven terms is $a+ar+ar^2+ar^3+ar^4+ar^5+ar^6=6+6\sqrt{3}+18+18\sqrt{3}+54+54\sqrt{3}+162=\boxed{306+78\sqrt{3}}$.