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Sagot :
To find the common ratio [tex]\(r\)[/tex] of a geometric progression (G.P.) where the sum of the first eight terms is five times the sum of the first four terms, we can follow these steps:
1. Express the sums using the sum formula for a G.P.
The sum of the first [tex]\(n\)[/tex] terms of a geometric progression is given by:
[tex]\[ S_n = a \frac{1 - r^n}{1 - r} \][/tex]
where [tex]\(a\)[/tex] is the first term, [tex]\(r\)[/tex] is the common ratio, and [tex]\(n\)[/tex] is the number of terms.
Therefore, the sum of the first eight terms ([tex]\( S_8 \)[/tex]) and the sum of the first four terms ([tex]\( S_4 \)[/tex]) can be written as:
[tex]\[ S_8 = a \frac{1 - r^8}{1 - r} \][/tex]
[tex]\[ S_4 = a \frac{1 - r^4}{1 - r} \][/tex]
2. Set up the equation according to the given condition:
We are given that the sum of the first eight terms is five times the sum of the first four terms:
[tex]\[ S_8 = 5 \cdot S_4 \][/tex]
Substituting the expressions for [tex]\( S_8 \)[/tex] and [tex]\( S_4 \)[/tex]:
[tex]\[ a \frac{1 - r^8}{1 - r} = 5 \left(a \frac{1 - r^4}{1 - r}\right) \][/tex]
3. Simplify the equation:
Since [tex]\( a \)[/tex] and [tex]\( (1 - r) \)[/tex] are common factors on both sides of the equation, they cancel out:
[tex]\[ 1 - r^8 = 5 (1 - r^4) \][/tex]
Simplifying further:
[tex]\[ 1 - r^8 = 5 - 5r^4 \][/tex]
Bring all the terms to one side to form a polynomial equation:
[tex]\[ r^8 - 5r^4 + 4 = 0 \][/tex]
4. Solve the polynomial equation by substitution:
Let [tex]\( x = r^4 \)[/tex]. This transforms the equation into a quadratic equation in terms of [tex]\( x \)[/tex]:
[tex]\[ x^2 - 5x + 4 = 0 \][/tex]
Solve for [tex]\( x \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 4 \)[/tex]:
[tex]\[ x = \frac{5 \pm \sqrt{25 - 16}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{9}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm 3}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = 4 \quad \text{and} \quad x = 1 \][/tex]
5. Relate [tex]\( x \)[/tex] back to [tex]\( r^4 \)[/tex]:
Recall that [tex]\( x = r^4 \)[/tex].
[tex]\[ r^4 = 4 \implies r = \pm \sqrt[4]{4} = \pm \sqrt{2} \][/tex]
[tex]\[ r^4 = 1 \implies r = \pm \sqrt[4]{1} = \pm 1 \][/tex]
6. Verify the valid values of [tex]\( r \)[/tex]:
Substitute [tex]\( r = \pm 1 \)[/tex] back into the original condition [tex]\( 1 - r^8 = 5 - 5r^4 \)[/tex]. We see that:
[tex]\[ 1 - 1^8 = 5 - 5 \cdot 1 \implies 0 \neq 0 \quad (\text{This does not hold true}) \][/tex]
Therefore, [tex]\( r = \pm 1 \)[/tex] are invalid solutions.
The only valid solutions for the common ratio [tex]\( r \)[/tex] are [tex]\( \sqrt{2} \)[/tex] and [tex]\( -\sqrt{2} \)[/tex]. Therefore, the correct answers are:
(a) [tex]\( \sqrt{2} \)[/tex]
(b) [tex]\( -\sqrt{2} \)[/tex]
Hence, the correct answer is:
(c) both
1. Express the sums using the sum formula for a G.P.
The sum of the first [tex]\(n\)[/tex] terms of a geometric progression is given by:
[tex]\[ S_n = a \frac{1 - r^n}{1 - r} \][/tex]
where [tex]\(a\)[/tex] is the first term, [tex]\(r\)[/tex] is the common ratio, and [tex]\(n\)[/tex] is the number of terms.
Therefore, the sum of the first eight terms ([tex]\( S_8 \)[/tex]) and the sum of the first four terms ([tex]\( S_4 \)[/tex]) can be written as:
[tex]\[ S_8 = a \frac{1 - r^8}{1 - r} \][/tex]
[tex]\[ S_4 = a \frac{1 - r^4}{1 - r} \][/tex]
2. Set up the equation according to the given condition:
We are given that the sum of the first eight terms is five times the sum of the first four terms:
[tex]\[ S_8 = 5 \cdot S_4 \][/tex]
Substituting the expressions for [tex]\( S_8 \)[/tex] and [tex]\( S_4 \)[/tex]:
[tex]\[ a \frac{1 - r^8}{1 - r} = 5 \left(a \frac{1 - r^4}{1 - r}\right) \][/tex]
3. Simplify the equation:
Since [tex]\( a \)[/tex] and [tex]\( (1 - r) \)[/tex] are common factors on both sides of the equation, they cancel out:
[tex]\[ 1 - r^8 = 5 (1 - r^4) \][/tex]
Simplifying further:
[tex]\[ 1 - r^8 = 5 - 5r^4 \][/tex]
Bring all the terms to one side to form a polynomial equation:
[tex]\[ r^8 - 5r^4 + 4 = 0 \][/tex]
4. Solve the polynomial equation by substitution:
Let [tex]\( x = r^4 \)[/tex]. This transforms the equation into a quadratic equation in terms of [tex]\( x \)[/tex]:
[tex]\[ x^2 - 5x + 4 = 0 \][/tex]
Solve for [tex]\( x \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 4 \)[/tex]:
[tex]\[ x = \frac{5 \pm \sqrt{25 - 16}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{9}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm 3}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = 4 \quad \text{and} \quad x = 1 \][/tex]
5. Relate [tex]\( x \)[/tex] back to [tex]\( r^4 \)[/tex]:
Recall that [tex]\( x = r^4 \)[/tex].
[tex]\[ r^4 = 4 \implies r = \pm \sqrt[4]{4} = \pm \sqrt{2} \][/tex]
[tex]\[ r^4 = 1 \implies r = \pm \sqrt[4]{1} = \pm 1 \][/tex]
6. Verify the valid values of [tex]\( r \)[/tex]:
Substitute [tex]\( r = \pm 1 \)[/tex] back into the original condition [tex]\( 1 - r^8 = 5 - 5r^4 \)[/tex]. We see that:
[tex]\[ 1 - 1^8 = 5 - 5 \cdot 1 \implies 0 \neq 0 \quad (\text{This does not hold true}) \][/tex]
Therefore, [tex]\( r = \pm 1 \)[/tex] are invalid solutions.
The only valid solutions for the common ratio [tex]\( r \)[/tex] are [tex]\( \sqrt{2} \)[/tex] and [tex]\( -\sqrt{2} \)[/tex]. Therefore, the correct answers are:
(a) [tex]\( \sqrt{2} \)[/tex]
(b) [tex]\( -\sqrt{2} \)[/tex]
Hence, the correct answer is:
(c) both
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