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Sagot :
Certainly! Let's solve the problem step-by-step:
### Step 1: Understanding the Geometric Series (G.S.)
In a geometric series, each term after the first is found by multiplying the previous term by a constant called the common ratio, denoted as [tex]\( r \)[/tex]. The [tex]\( n \)[/tex]-th term [tex]\( T_n \)[/tex] of a geometric series can be represented as:
[tex]\[ T_n = a \cdot r^{n-1} \][/tex]
where [tex]\( a \)[/tex] is the first term.
### Step 2: Representing Given Terms in the Formula
We are given:
- The 3rd term ([tex]\( T_3 \)[/tex]) is 27:
[tex]\[ T_3 = a \cdot r^2 = 27 \][/tex]
- The 5th term ([tex]\( T_5 \)[/tex]) is 3:
[tex]\[ T_5 = a \cdot r^4 = 3 \][/tex]
### Step 3: Finding the Common Ratio [tex]\( r \)[/tex]
To find the common ratio [tex]\( r \)[/tex], we can divide the equation for the 5th term by the equation for the 3rd term:
[tex]\[ \frac{a \cdot r^4}{a \cdot r^2} = \frac{3}{27} \][/tex]
Simplifying this, we get:
[tex]\[ r^2 = \frac{1}{9} \][/tex]
[tex]\[ r = \frac{1}{3} \][/tex]
(Note: We take the positive value for [tex]\( r \)[/tex] as we're dealing with a standard geometric series.)
### Step 4: Finding the First Term [tex]\( a \)[/tex]
With [tex]\( r \)[/tex] found, we substitute it back into the equation for the 3rd term to find [tex]\( a \)[/tex]:
[tex]\[ a \cdot \left(\frac{1}{3}\right)^2 = 27 \][/tex]
[tex]\[ a \cdot \frac{1}{9} = 27 \][/tex]
[tex]\[ a = 27 \cdot 9 \][/tex]
[tex]\[ a = 243 \][/tex]
### Step 5: Find the Term Which is [tex]\( \frac{1}{9} \)[/tex]
We want to find the term [tex]\( T_n \)[/tex] which equals [tex]\( \frac{1}{9} \)[/tex]. Using the formula for the [tex]\( n \)[/tex]-th term:
[tex]\[ a \cdot r^{n-1} = \frac{1}{9} \][/tex]
Substitute [tex]\( a = 243 \)[/tex] and [tex]\( r = \frac{1}{3} \)[/tex] into the equation:
[tex]\[ 243 \cdot \left(\frac{1}{3}\right)^{n-1} = \frac{1}{9} \][/tex]
### Step 6: Simplifying the Equation
[tex]\[ \left(\frac{1}{3}\right)^{n-1} = \frac{1}{2187} \][/tex]
Since [tex]\( 243 \cdot 9 = 2187 \)[/tex], and knowing [tex]\( (1/3)^7 = 1/2187 \)[/tex]:
[tex]\[ (1/3)^{n-1} = (1/3)^7 \][/tex]
Thus:
[tex]\[ n-1 = 7 \][/tex]
[tex]\[ n = 8 \][/tex]
### Conclusion
Therefore, the term in the geometric series that equals [tex]\( \frac{1}{9} \)[/tex] is the 8th term.
### Step 1: Understanding the Geometric Series (G.S.)
In a geometric series, each term after the first is found by multiplying the previous term by a constant called the common ratio, denoted as [tex]\( r \)[/tex]. The [tex]\( n \)[/tex]-th term [tex]\( T_n \)[/tex] of a geometric series can be represented as:
[tex]\[ T_n = a \cdot r^{n-1} \][/tex]
where [tex]\( a \)[/tex] is the first term.
### Step 2: Representing Given Terms in the Formula
We are given:
- The 3rd term ([tex]\( T_3 \)[/tex]) is 27:
[tex]\[ T_3 = a \cdot r^2 = 27 \][/tex]
- The 5th term ([tex]\( T_5 \)[/tex]) is 3:
[tex]\[ T_5 = a \cdot r^4 = 3 \][/tex]
### Step 3: Finding the Common Ratio [tex]\( r \)[/tex]
To find the common ratio [tex]\( r \)[/tex], we can divide the equation for the 5th term by the equation for the 3rd term:
[tex]\[ \frac{a \cdot r^4}{a \cdot r^2} = \frac{3}{27} \][/tex]
Simplifying this, we get:
[tex]\[ r^2 = \frac{1}{9} \][/tex]
[tex]\[ r = \frac{1}{3} \][/tex]
(Note: We take the positive value for [tex]\( r \)[/tex] as we're dealing with a standard geometric series.)
### Step 4: Finding the First Term [tex]\( a \)[/tex]
With [tex]\( r \)[/tex] found, we substitute it back into the equation for the 3rd term to find [tex]\( a \)[/tex]:
[tex]\[ a \cdot \left(\frac{1}{3}\right)^2 = 27 \][/tex]
[tex]\[ a \cdot \frac{1}{9} = 27 \][/tex]
[tex]\[ a = 27 \cdot 9 \][/tex]
[tex]\[ a = 243 \][/tex]
### Step 5: Find the Term Which is [tex]\( \frac{1}{9} \)[/tex]
We want to find the term [tex]\( T_n \)[/tex] which equals [tex]\( \frac{1}{9} \)[/tex]. Using the formula for the [tex]\( n \)[/tex]-th term:
[tex]\[ a \cdot r^{n-1} = \frac{1}{9} \][/tex]
Substitute [tex]\( a = 243 \)[/tex] and [tex]\( r = \frac{1}{3} \)[/tex] into the equation:
[tex]\[ 243 \cdot \left(\frac{1}{3}\right)^{n-1} = \frac{1}{9} \][/tex]
### Step 6: Simplifying the Equation
[tex]\[ \left(\frac{1}{3}\right)^{n-1} = \frac{1}{2187} \][/tex]
Since [tex]\( 243 \cdot 9 = 2187 \)[/tex], and knowing [tex]\( (1/3)^7 = 1/2187 \)[/tex]:
[tex]\[ (1/3)^{n-1} = (1/3)^7 \][/tex]
Thus:
[tex]\[ n-1 = 7 \][/tex]
[tex]\[ n = 8 \][/tex]
### Conclusion
Therefore, the term in the geometric series that equals [tex]\( \frac{1}{9} \)[/tex] is the 8th term.
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