Answer :
Sure! Let's consider the problem step-by-step to identify the term in the given geometric series that equals 32.
Firstly, identify the key components of the geometric series:
- The first term [tex]\(a\)[/tex] of the series is [tex]\(\frac{1}{8}\)[/tex].
- The common ratio [tex]\(r\)[/tex] is obtained by dividing any term by its preceding term. Taking the second term [tex]\(\frac{1}{4}\)[/tex] and dividing by the first term [tex]\(\frac{1}{8}\)[/tex], we get:
[tex]\[ r = \frac{\frac{1}{4}}{\frac{1}{8}} = \frac{1}{4} \times \frac{8}{1} = 2 \][/tex]
So the series is:
[tex]\[ \frac{1}{8}, \frac{1}{4}, \frac{1}{2}, \ldots, \ldots, 32 \][/tex]
A geometric series is defined with the [tex]\(n\)[/tex]-th term [tex]\(T_n\)[/tex] given by the formula:
[tex]\[ T_n = a \cdot r^{(n-1)} \][/tex]
We need to find [tex]\(n\)[/tex] such that [tex]\(T_n = 32\)[/tex]. So, set up the equation:
[tex]\[ 32 = \frac{1}{8} \cdot 2^{(n-1)} \][/tex]
Now solve for [tex]\(n\)[/tex]:
1. Multiply both sides of the equation by 8 to isolate the exponential term:
[tex]\[ 256 = 2^{(n-1)} \][/tex]
2. Recall that 256 is a power of 2 (since [tex]\(2^8 = 256\)[/tex]):
[tex]\[ 256 = 2^8 \][/tex]
So:
[tex]\[ 2^{(n-1)} = 2^8 \][/tex]
3. Since the bases are the same, the exponents must be equal:
[tex]\[ n-1 = 8 \][/tex]
4. Solving this equation for [tex]\(n\)[/tex]:
[tex]\[ n = 8 + 1 = 9 \][/tex]
Thus, the term in the series [tex]\(\frac{1}{8}+\frac{1}{4}+\frac{1}{2}+\ldots+32\)[/tex] that is equal to 32 is the 9th term.
Firstly, identify the key components of the geometric series:
- The first term [tex]\(a\)[/tex] of the series is [tex]\(\frac{1}{8}\)[/tex].
- The common ratio [tex]\(r\)[/tex] is obtained by dividing any term by its preceding term. Taking the second term [tex]\(\frac{1}{4}\)[/tex] and dividing by the first term [tex]\(\frac{1}{8}\)[/tex], we get:
[tex]\[ r = \frac{\frac{1}{4}}{\frac{1}{8}} = \frac{1}{4} \times \frac{8}{1} = 2 \][/tex]
So the series is:
[tex]\[ \frac{1}{8}, \frac{1}{4}, \frac{1}{2}, \ldots, \ldots, 32 \][/tex]
A geometric series is defined with the [tex]\(n\)[/tex]-th term [tex]\(T_n\)[/tex] given by the formula:
[tex]\[ T_n = a \cdot r^{(n-1)} \][/tex]
We need to find [tex]\(n\)[/tex] such that [tex]\(T_n = 32\)[/tex]. So, set up the equation:
[tex]\[ 32 = \frac{1}{8} \cdot 2^{(n-1)} \][/tex]
Now solve for [tex]\(n\)[/tex]:
1. Multiply both sides of the equation by 8 to isolate the exponential term:
[tex]\[ 256 = 2^{(n-1)} \][/tex]
2. Recall that 256 is a power of 2 (since [tex]\(2^8 = 256\)[/tex]):
[tex]\[ 256 = 2^8 \][/tex]
So:
[tex]\[ 2^{(n-1)} = 2^8 \][/tex]
3. Since the bases are the same, the exponents must be equal:
[tex]\[ n-1 = 8 \][/tex]
4. Solving this equation for [tex]\(n\)[/tex]:
[tex]\[ n = 8 + 1 = 9 \][/tex]
Thus, the term in the series [tex]\(\frac{1}{8}+\frac{1}{4}+\frac{1}{2}+\ldots+32\)[/tex] that is equal to 32 is the 9th term.
