Respuesta :
The three numbers are 4 , 8 , 16
Step-by-step explanation:
Let us revise how to find the nth term of the geometric and the arithmetic progressions
- The nth term of the geometric sequence is [tex]a_{n}=ar^{n-1}[/tex] , where a is the first term and r is the common ratio between the consecutive terms
- The nth term of the arithmetic sequence is [tex]a_{n}=a+(n-1)d[/tex] , where a is the first term and d in the common difference between the consecutive terms
∵ Three numbers form a geometric progression
- Assume that the first number is a and the common ratio is r
and n = 1 , 2 , 3
∴ The three terms are a , ar and ar²
∵ The second term is increased by 2
∴ The three terms are a , ar + 2 , ar²
∵ The three terms formed an arithmetic progression
- The common difference between each two consecutive terms is d
∴ d = ar + 2 - a and d = ar² - (ar + 2)
- Equate the right hand sides of d
∴ ar + 2 - a = ar² - ar - 2
- Add 2 to both sides
∴ ar - a + 4 = ar² - ar
- Subtract ar from both sides
∴ -a + 4 = ar² - 2ar
- Add a to both sides
∴ 4 = ar² - 2ar + a
- Take a as a common factor in the right hand side
∴ 4 = a(r² - 2r + 1)
∵ r² - 2r + 1 = (r - 1)²
∴ 4 = a(r - 1)²
- Divide both sides by (r - 1)²
∴ [tex]a=\frac{4}{(r-1)^{2}}[/tex] ⇒ (1)
∵ The last term is increased by 9
∴ The three terms are a , ar + 2 , ar² + 9
∵ The three terms formed an geometric progression
- Find the common ratio between each two consecutive terms
∵ The common ratio = [tex]\frac{ar+2}{a}[/tex]
∵ The common ratio = [tex]\frac{ar^{2}+9}{ar+2}[/tex]
- Equate the right hand sides of the common ratio
∴ [tex]\frac{ar+2}{a}=\frac{ar^{2}+9}{ar+2}[/tex]
- By using cross multiplication
∴ a(ar² + 9) = (ar + 2)²
- Simplify the two sides
∴ a²r² + 9a = a²r² + 4ar + 4
- Subtract a²r² from both sides
∴ 9a = 4ar + 4
- Subtract 4ar from both sides
∴ 9a - 4ar = 4
- Take a as a common factor from both sides
∴ a(9 - 4r) = 4
- Divide both sides by (9 - 4r)
∴ [tex]a=\frac{4}{(9-4r)}[/tex] ⇒ (2)
Equate the right hand sides of (1) and (2)
∴ [tex]\frac{4}{(r-1)^{2}}[/tex] = [tex]\frac{4}{(9-4r)}[/tex]
- By using cross multiplication
∴ 4(r - 1)² = 4(9 - 4r)
- Divide both sides by 4
∴ (r - 1)² = 9 - 4r
- Solve the bracket of the left hand side
∴ r² - 2r + 1 = 9 - 4r
- Add 4r to both sides
∴ r² + 2r + 1 = 9
- Subtract 9 from both sides
∴ r² + 2r - 8 = 0
- Factorize it into 2 factors
∴ (r - 2)(r + 4) = 0
- Equate each factor by 0
∵ r - 2 = 0
- Add 2 to both sides
∴ r = 2
∵ r + 4 = 0
- Subtract 4 from both sides
∴ r = -4 ⇒ rejected
Substitute the value of r in equation (2) to find a
∵ [tex]a=\frac{4}{9-4(2)}=\frac{4}{9-8}=\frac{4}{1}[/tex]
∴ a = 4
∵ The three numbers are a , ar , ar²
∵ a = 4 and r = 2
∴ The numbers are 4 , 4(2) , 4(2)²
∴ The numbers are 4 , 8 , 16
Lets check the answer
4 , 8 + 2 , 16 ⇒ 4 , 10 , 16 formed an arithmetic progression with common difference 6
4 , 10 , 16 + 9 ⇒ 4 , 10 , 25 formed a geometric progression with common ratio 2.5
The three numbers are 4 , 8 , 16
Learn more:
You can learn more about the progressions in brainly.com/question/1522572
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The three terms in geometric progression are 4, 8, and 16 and this can be determined by using the properties of geometric progression.
Given :
- Three numbers form a geometric progression.
- If the second term is increased by 2, then the progression will become arithmetic.
- If, after this, the last term is increased by 9, then the progression will again become geometric.
Let the three terms in geometric progression be a, ar, and [tex]ar^2[/tex]. Given that if the second term is increased by 2 then the progression will become arithmetic that is:
[tex]\rm d =ar+2-a[/tex] and [tex]\rm d = ar^2 - (ar+2)[/tex]
Now, equate both the equation.
[tex]ar+2-a=ar^2-ar-2[/tex]
[tex]2ar+4=ar^2+a[/tex]
[tex]4 = a(r^2-2r+1)[/tex]
Factorize the above equation.
[tex]4= a(r-1)^2[/tex]
[tex]a = \dfrac{4}{(r-1)^2}[/tex] ---- (1)
Now, it is also given that the last term is increased by 9, that is:
a, ar + 2, a[tex]\rm r^2[/tex] + 9
The common ratio of the above geometric progression is:
[tex]\rm Common \; Ratio = \dfrac{ar+2}{a}[/tex] ---- (2)
[tex]\rm Common \; Ratio = \dfrac{ar^2+9}{ar+2}[/tex] ---- (3)
Now, equate both the equation.
[tex]\rm \dfrac{ar+2}{a}=\dfrac{ar^2+9}{ar+2}[/tex]
Cross multiply in the above equation.
[tex]\rm (ar+2)^2=a(ar^2+9)[/tex]
[tex]\rm a^2r^2+4+4ar=a^2r^2+9a[/tex]
9a - 4ar - 4 = 0
[tex]\rm a = \dfrac{4}{9-4r}[/tex] --- (4)
Now, equate equation (1) and equation (4).
[tex]\dfrac{4}{9-4r} = \dfrac{4}{(r-1)^2}[/tex]
Cross multiply in the above equation.
[tex]\rm r^2-2r+1 = 9-4r[/tex]
[tex]\rm r^2+2r-8=0[/tex]
Factorize the above equation.
[tex]\rm r^2 +4r - 2r -8=0[/tex]
(r + 4)(r - 2) = 0
r = 2 (Accepted)
Now, put the value of r in the equation (4).
[tex]a =\dfrac{4}{9-4(2)}[/tex]
a = 4
ar = 4(2) = 8
a[tex]r^2[/tex] = 4(4) = 16
The three terms in geometric progression are 4, 8, and 16.
For more information, refer to the link given below:
https://brainly.com/question/22687297
