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Sagot :
To find the greatest possible value for [tex]\( t_5 \)[/tex] in a geometric series where the first term [tex]\( t_1 = 12 \)[/tex] and the sum of the first three terms [tex]\( S_3 = 372 \)[/tex], follow these steps:
1. Express the Sum of the First Three Terms:
The sum of the first three terms of a geometric series is given by:
[tex]\[ S_3 = t_1 \left( \frac{1 - r^3}{1 - r} \right) \][/tex]
Given [tex]\( S_3 = 372 \)[/tex] and [tex]\( t_1 = 12 \)[/tex], substitute these values into the equation:
[tex]\[ 372 = 12 \left( \frac{1 - r^3}{1 - r} \right) \][/tex]
2. Simplify the Equation:
Divide both sides of the equation by 12:
[tex]\[ 31 = \frac{1 - r^3}{1 - r} \][/tex]
Notice that [tex]\( \frac{1 - r^3}{1 - r} \)[/tex] is a well-known formula for the sum of a geometric series up to [tex]\( r^2 \)[/tex], which simplifies to:
[tex]\[ 1 + r + r^2 \][/tex]
Therefore, we have:
[tex]\[ 31 = 1 + r + r^2 \][/tex]
3. Form a Quadratic Equation:
Rearrange the equation into a standard quadratic form:
[tex]\[ r^2 + r - 30 = 0 \][/tex]
4. Solve the Quadratic Equation:
Use the quadratic formula:
[tex]\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -30 \)[/tex]. Substitute these values into the formula:
[tex]\[ r = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1} \][/tex]
[tex]\[ r = \frac{-1 \pm \sqrt{1 + 120}}{2} \][/tex]
[tex]\[ r = \frac{-1 \pm \sqrt{121}}{2} \][/tex]
[tex]\[ r = \frac{-1 \pm 11}{2} \][/tex]
Thus, we get two roots:
[tex]\[ r_1 = \frac{-1 + 11}{2} = 5 \][/tex]
[tex]\[ r_2 = \frac{-1 - 11}{2} = -6 \][/tex]
5. Calculate the Fifth Term [tex]\( t_5 \)[/tex]:
The nth term of a geometric series is given by:
[tex]\[ t_n = t_1 \cdot r^{n-1} \][/tex]
For [tex]\( t_5 \)[/tex]:
[tex]\[ t_5 = t_1 \cdot r^4 \][/tex]
Substitute [tex]\( t_1 = 12 \)[/tex] and the two values of [tex]\( r \)[/tex]:
- For [tex]\( r_1 = 5 \)[/tex]:
[tex]\[ t_5 = 12 \cdot 5^4 = 12 \cdot 625 = 7500 \][/tex]
- For [tex]\( r_2 = -6 \)[/tex]:
[tex]\[ t_5 = 12 \cdot (-6)^4 = 12 \cdot 1296 = 15552 \][/tex]
6. Identify the Greatest Possible Value:
Compare the two values of [tex]\( t_5 \)[/tex]:
[tex]\[ \text{Greatest possible value of } t_5 = \max(7500, 15552) = 15552 \][/tex]
Therefore, the greatest possible value for [tex]\( t_5 \)[/tex] is [tex]\( 15552 \)[/tex].
1. Express the Sum of the First Three Terms:
The sum of the first three terms of a geometric series is given by:
[tex]\[ S_3 = t_1 \left( \frac{1 - r^3}{1 - r} \right) \][/tex]
Given [tex]\( S_3 = 372 \)[/tex] and [tex]\( t_1 = 12 \)[/tex], substitute these values into the equation:
[tex]\[ 372 = 12 \left( \frac{1 - r^3}{1 - r} \right) \][/tex]
2. Simplify the Equation:
Divide both sides of the equation by 12:
[tex]\[ 31 = \frac{1 - r^3}{1 - r} \][/tex]
Notice that [tex]\( \frac{1 - r^3}{1 - r} \)[/tex] is a well-known formula for the sum of a geometric series up to [tex]\( r^2 \)[/tex], which simplifies to:
[tex]\[ 1 + r + r^2 \][/tex]
Therefore, we have:
[tex]\[ 31 = 1 + r + r^2 \][/tex]
3. Form a Quadratic Equation:
Rearrange the equation into a standard quadratic form:
[tex]\[ r^2 + r - 30 = 0 \][/tex]
4. Solve the Quadratic Equation:
Use the quadratic formula:
[tex]\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -30 \)[/tex]. Substitute these values into the formula:
[tex]\[ r = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1} \][/tex]
[tex]\[ r = \frac{-1 \pm \sqrt{1 + 120}}{2} \][/tex]
[tex]\[ r = \frac{-1 \pm \sqrt{121}}{2} \][/tex]
[tex]\[ r = \frac{-1 \pm 11}{2} \][/tex]
Thus, we get two roots:
[tex]\[ r_1 = \frac{-1 + 11}{2} = 5 \][/tex]
[tex]\[ r_2 = \frac{-1 - 11}{2} = -6 \][/tex]
5. Calculate the Fifth Term [tex]\( t_5 \)[/tex]:
The nth term of a geometric series is given by:
[tex]\[ t_n = t_1 \cdot r^{n-1} \][/tex]
For [tex]\( t_5 \)[/tex]:
[tex]\[ t_5 = t_1 \cdot r^4 \][/tex]
Substitute [tex]\( t_1 = 12 \)[/tex] and the two values of [tex]\( r \)[/tex]:
- For [tex]\( r_1 = 5 \)[/tex]:
[tex]\[ t_5 = 12 \cdot 5^4 = 12 \cdot 625 = 7500 \][/tex]
- For [tex]\( r_2 = -6 \)[/tex]:
[tex]\[ t_5 = 12 \cdot (-6)^4 = 12 \cdot 1296 = 15552 \][/tex]
6. Identify the Greatest Possible Value:
Compare the two values of [tex]\( t_5 \)[/tex]:
[tex]\[ \text{Greatest possible value of } t_5 = \max(7500, 15552) = 15552 \][/tex]
Therefore, the greatest possible value for [tex]\( t_5 \)[/tex] is [tex]\( 15552 \)[/tex].
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