Answer :
To find the sum of the first five terms of a geometric series with the first term [tex]\( a_1 = 6 \)[/tex] and common ratio [tex]\( r = \frac{1}{3} \)[/tex], we use the formula for the sum of the first [tex]\( n \)[/tex] terms of a geometric series:
[tex]\[ S_n = a_1 \frac{1 - r^n}{1 - r} \][/tex]
Where [tex]\( S_n \)[/tex] is the sum of the first [tex]\( n \)[/tex] terms, [tex]\( a_1 \)[/tex] is the first term, [tex]\( r \)[/tex] is the common ratio, and [tex]\( n \)[/tex] is the number of terms. Substitute [tex]\( a_1 = 6 \)[/tex], [tex]\( r = \frac{1}{3} \)[/tex], and [tex]\( n = 5 \)[/tex]:
[tex]\[ S_5 = 6 \frac{1 - \left(\frac{1}{3}\right)^5}{1 - \frac{1}{3}} \][/tex]
First, calculate [tex]\( \left(\frac{1}{3}\right)^5 \)[/tex]:
[tex]\[ \left(\frac{1}{3}\right)^5 = \frac{1}{243} \][/tex]
Then, substitute this value back into the formula:
[tex]\[ S_5 = 6 \frac{1 - \frac{1}{243}}{1 - \frac{1}{3}} \][/tex]
Simplify the numerator and the denominator:
[tex]\[ 1 - \frac{1}{243} = \frac{243}{243} - \frac{1}{243} = \frac{242}{243} \][/tex]
[tex]\[ 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3} \][/tex]
Now, our formula looks like this:
[tex]\[ S_5 = 6 \frac{\frac{242}{243}}{\frac{2}{3}} \][/tex]
To simplify the complex fraction, multiply by the reciprocal of the denominator:
[tex]\[ S_5 = 6 \left(\frac{242}{243} \times \frac{3}{2}\right) \][/tex]
Multiply the numerators and denominators:
[tex]\[ S_5 = 6 \frac{242 \times 3}{243 \times 2} = 6 \frac{726}{486} \][/tex]
Further simplification (since 726 and 486 are divisible by 6):
[tex]\[ \frac{726}{486} = \frac{121}{81} \][/tex]
Now, multiply by 6:
[tex]\[ S_5 = 6 \times \frac{121}{81} = \frac{726}{81} \][/tex]
Simplify by dividing both the numerator and denominator by 3:
[tex]\[ \frac{726}{81} = \frac{242}{27} \][/tex]
Thus, the sum of the first five terms of the geometric series is:
[tex]\[ \boxed{\frac{242}{27}} \][/tex]
[tex]\[ S_n = a_1 \frac{1 - r^n}{1 - r} \][/tex]
Where [tex]\( S_n \)[/tex] is the sum of the first [tex]\( n \)[/tex] terms, [tex]\( a_1 \)[/tex] is the first term, [tex]\( r \)[/tex] is the common ratio, and [tex]\( n \)[/tex] is the number of terms. Substitute [tex]\( a_1 = 6 \)[/tex], [tex]\( r = \frac{1}{3} \)[/tex], and [tex]\( n = 5 \)[/tex]:
[tex]\[ S_5 = 6 \frac{1 - \left(\frac{1}{3}\right)^5}{1 - \frac{1}{3}} \][/tex]
First, calculate [tex]\( \left(\frac{1}{3}\right)^5 \)[/tex]:
[tex]\[ \left(\frac{1}{3}\right)^5 = \frac{1}{243} \][/tex]
Then, substitute this value back into the formula:
[tex]\[ S_5 = 6 \frac{1 - \frac{1}{243}}{1 - \frac{1}{3}} \][/tex]
Simplify the numerator and the denominator:
[tex]\[ 1 - \frac{1}{243} = \frac{243}{243} - \frac{1}{243} = \frac{242}{243} \][/tex]
[tex]\[ 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3} \][/tex]
Now, our formula looks like this:
[tex]\[ S_5 = 6 \frac{\frac{242}{243}}{\frac{2}{3}} \][/tex]
To simplify the complex fraction, multiply by the reciprocal of the denominator:
[tex]\[ S_5 = 6 \left(\frac{242}{243} \times \frac{3}{2}\right) \][/tex]
Multiply the numerators and denominators:
[tex]\[ S_5 = 6 \frac{242 \times 3}{243 \times 2} = 6 \frac{726}{486} \][/tex]
Further simplification (since 726 and 486 are divisible by 6):
[tex]\[ \frac{726}{486} = \frac{121}{81} \][/tex]
Now, multiply by 6:
[tex]\[ S_5 = 6 \times \frac{121}{81} = \frac{726}{81} \][/tex]
Simplify by dividing both the numerator and denominator by 3:
[tex]\[ \frac{726}{81} = \frac{242}{27} \][/tex]
Thus, the sum of the first five terms of the geometric series is:
[tex]\[ \boxed{\frac{242}{27}} \][/tex]