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Sagot :
Let's solve this step-by-step.
Given:
1. The sum of the first three terms of the geometric sequence is 26.
2. The product of the first three terms is 216.
We need to find the common ratio [tex]\( r \)[/tex].
### Step 1: Define the terms
Let the first term be [tex]\( a \)[/tex]. Then the first three terms of the geometric sequence can be written as:
- First term: [tex]\( a \)[/tex]
- Second term: [tex]\( ar \)[/tex]
- Third term: [tex]\( ar^2 \)[/tex]
### Step 2: Set up the equations
From the given information, we can set up the following two equations:
1. Sum of the first three terms:
[tex]\[ a + ar + ar^2 = 26 \][/tex]
2. Product of the first three terms:
[tex]\[ a \times ar \times ar^2 = 216 \][/tex]
### Step 3: Simplify the product equation
The product equation can be simplified as:
[tex]\[ a \cdot ar \cdot ar^2 = a^3 r^3 = 216 \][/tex]
[tex]\[ (a r)^{3} = 216 \][/tex]
Taking the cube root of both sides:
[tex]\[ a r = \sqrt[3]{216} \][/tex]
[tex]\[ a r = 6 \][/tex]
This gives us the relationship:
[tex]\[ a = \frac{6}{r} \][/tex]
### Step 4: Substitute [tex]\( a \)[/tex] into the sum equation
Using [tex]\( a = \frac{6}{r} \)[/tex] in the sum equation:
[tex]\[ \frac{6}{r} + \frac{6}{r} \cdot r + \frac{6}{r} \cdot r^2 = 26 \][/tex]
[tex]\[ \frac{6}{r} + 6 + 6r = 26 \][/tex]
Clear the fraction by multiplying through by [tex]\( r \)[/tex]:
[tex]\[ 6 + 6r + 6r^2 = 26r \][/tex]
### Step 5: Rearrange and solve the quadratic equation
[tex]\[ 6r^2 + 6r + 6 = 26r \][/tex]
[tex]\[ 6r^2 + 6r + 6 - 26r = 0 \][/tex]
[tex]\[ 6r^2 - 20r + 6 = 0 \][/tex]
Divide through by 2 to simplify:
[tex]\[ 3r^2 - 10r + 3 = 0 \][/tex]
### Step 6: Solve the quadratic equation
We will solve [tex]\( 3r^2 - 10r + 3 = 0 \)[/tex] using the quadratic formula [tex]\( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = 3 \)[/tex].
Calculating the discriminant:
[tex]\[ b^2 - 4ac = (-10)^2 - 4 \cdot 3 \cdot 3 = 100 - 36 = 64 \][/tex]
So,
[tex]\[ r = \frac{10 \pm \sqrt{64}}{6} \][/tex]
[tex]\[ r = \frac{10 \pm 8}{6} \][/tex]
This yields two solutions for [tex]\( r \)[/tex]:
[tex]\[ r = \frac{10 + 8}{6} = \frac{18}{6} = 3 \][/tex]
[tex]\[ r = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3} \][/tex]
### Step 7: Verification
We should verify both [tex]\( r = 3 \)[/tex] and [tex]\( r = \frac{1}{3} \)[/tex].
For [tex]\( r = 3 \)[/tex]:
[tex]\[ a = \frac{6}{3} = 2 \][/tex]
First three terms: 2, 6, 18
Sum: [tex]\( 2 + 6 + 18 = 26 \)[/tex]
Product: [tex]\( 2 \times 6 \times 18 = 216 \)[/tex]
For [tex]\( r = \frac{1}{3} \)[/tex]:
[tex]\[ a = \frac{6}{\frac{1}{3}} = 18 \][/tex]
First three terms: 18, 6, 2
Sum: [tex]\( 18 + 6 + 2 = 26 \)[/tex]
Product: [tex]\( 18 \times 6 \times 2 = 216 \)[/tex]
Both solutions are valid. Hence,
[tex]\[ r = 3 \quad \text{or} \quad r = \frac{1}{3} \][/tex]
Thus, the common ratio can be either [tex]\( 3 \)[/tex] or [tex]\( \frac{1}{3} \)[/tex].
Given:
1. The sum of the first three terms of the geometric sequence is 26.
2. The product of the first three terms is 216.
We need to find the common ratio [tex]\( r \)[/tex].
### Step 1: Define the terms
Let the first term be [tex]\( a \)[/tex]. Then the first three terms of the geometric sequence can be written as:
- First term: [tex]\( a \)[/tex]
- Second term: [tex]\( ar \)[/tex]
- Third term: [tex]\( ar^2 \)[/tex]
### Step 2: Set up the equations
From the given information, we can set up the following two equations:
1. Sum of the first three terms:
[tex]\[ a + ar + ar^2 = 26 \][/tex]
2. Product of the first three terms:
[tex]\[ a \times ar \times ar^2 = 216 \][/tex]
### Step 3: Simplify the product equation
The product equation can be simplified as:
[tex]\[ a \cdot ar \cdot ar^2 = a^3 r^3 = 216 \][/tex]
[tex]\[ (a r)^{3} = 216 \][/tex]
Taking the cube root of both sides:
[tex]\[ a r = \sqrt[3]{216} \][/tex]
[tex]\[ a r = 6 \][/tex]
This gives us the relationship:
[tex]\[ a = \frac{6}{r} \][/tex]
### Step 4: Substitute [tex]\( a \)[/tex] into the sum equation
Using [tex]\( a = \frac{6}{r} \)[/tex] in the sum equation:
[tex]\[ \frac{6}{r} + \frac{6}{r} \cdot r + \frac{6}{r} \cdot r^2 = 26 \][/tex]
[tex]\[ \frac{6}{r} + 6 + 6r = 26 \][/tex]
Clear the fraction by multiplying through by [tex]\( r \)[/tex]:
[tex]\[ 6 + 6r + 6r^2 = 26r \][/tex]
### Step 5: Rearrange and solve the quadratic equation
[tex]\[ 6r^2 + 6r + 6 = 26r \][/tex]
[tex]\[ 6r^2 + 6r + 6 - 26r = 0 \][/tex]
[tex]\[ 6r^2 - 20r + 6 = 0 \][/tex]
Divide through by 2 to simplify:
[tex]\[ 3r^2 - 10r + 3 = 0 \][/tex]
### Step 6: Solve the quadratic equation
We will solve [tex]\( 3r^2 - 10r + 3 = 0 \)[/tex] using the quadratic formula [tex]\( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = 3 \)[/tex].
Calculating the discriminant:
[tex]\[ b^2 - 4ac = (-10)^2 - 4 \cdot 3 \cdot 3 = 100 - 36 = 64 \][/tex]
So,
[tex]\[ r = \frac{10 \pm \sqrt{64}}{6} \][/tex]
[tex]\[ r = \frac{10 \pm 8}{6} \][/tex]
This yields two solutions for [tex]\( r \)[/tex]:
[tex]\[ r = \frac{10 + 8}{6} = \frac{18}{6} = 3 \][/tex]
[tex]\[ r = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3} \][/tex]
### Step 7: Verification
We should verify both [tex]\( r = 3 \)[/tex] and [tex]\( r = \frac{1}{3} \)[/tex].
For [tex]\( r = 3 \)[/tex]:
[tex]\[ a = \frac{6}{3} = 2 \][/tex]
First three terms: 2, 6, 18
Sum: [tex]\( 2 + 6 + 18 = 26 \)[/tex]
Product: [tex]\( 2 \times 6 \times 18 = 216 \)[/tex]
For [tex]\( r = \frac{1}{3} \)[/tex]:
[tex]\[ a = \frac{6}{\frac{1}{3}} = 18 \][/tex]
First three terms: 18, 6, 2
Sum: [tex]\( 18 + 6 + 2 = 26 \)[/tex]
Product: [tex]\( 18 \times 6 \times 2 = 216 \)[/tex]
Both solutions are valid. Hence,
[tex]\[ r = 3 \quad \text{or} \quad r = \frac{1}{3} \][/tex]
Thus, the common ratio can be either [tex]\( 3 \)[/tex] or [tex]\( \frac{1}{3} \)[/tex].
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