Question

Three numbers form a geometric progression. If the second term is increased by 2, then the progression will become arithmetic and if, after this, the last term is increased by 9, then the progression will again become geometric. Find these three numbers.

Answer 1

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 $a_{n}=ar^{n-1}$ , where a is the first term and r is the common ratio between the consecutive terms
• The nth term of the arithmetic sequence is $a_{n}=a+(n-1)d$ , 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

∴ $a=\frac{4}{(r-1)^{2}}$ ⇒ (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 = $\frac{ar+2}{a}$

∵ The common ratio = $\frac{ar^{2}+9}{ar+2}$

- Equate the right hand sides of the common ratio

∴ $\frac{ar+2}{a}=\frac{ar^{2}+9}{ar+2}$

- 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)

∴ $a=\frac{4}{(9-4r)}$ ⇒ (2)

Equate the right hand sides of (1) and (2)

∴ $\frac{4}{(r-1)^{2}}$ = $\frac{4}{(9-4r)}$

- 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

∵ $a=\frac{4}{9-4(2)}=\frac{4}{9-8}=\frac{4}{1}$

∴ 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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