If 2x+1,x square +x+1and 3x square-3x+3 are in A.P,then the value (s) of x:
Answers
Answer :
The value of x is 1 (or) 2
Step-by-step explanation :
Arithmetic Progression :
- It is the sequence of numbers such that the difference between any two successive numbers is constant.
- General form of AP,
a , a+d , a+2d , a+3d , ..........
Given,
⇒ 2x + 1 , x² + x + 1 and 3x² - 3x + 3 are in A.P
Let
- a₁ = 2x + 1
- a₂ = x² + x + 1
- a₃ = 3x² - 3x + 3
Since they are in A.P., the difference between the successive numbers is constant.
a₂ - a₁ = a₃ - a₂
x² + x + 1 - (2x + 1) = 3x² - 3x + 3 - (x² + x + 1)
x² + x + 1 - 2x - 1 = 3x² - 3x + 3 - x² - x - 1
x² + x - 2x + 1 - 1 = 3x² - x² - 3x - x + 3 - 1
x² - x = 2x² - 4x + 2
2x² - x² + 2 = 4x - x
x² + 2 = 3x
x² - 3x + 2 = 0
x² - x - 2x + 2 = 0
x(x - 1) - 2(x - 1) = 0
(x - 1) (x - 2) = 0
x = 1 (or) 2
Therefore, the value of x is 1 (or) 2
Verification :
- Put x = 1,
⇒ a₁ = 2x + 1
a₁ = 2(1) + 1
a₁ = 2 + 1
a₁ = 3
⇒ a₂ = x² + x + 1
a₂ = 1² + 1 + 1
a₂ = 1 + 1 + 1
a₂ = 3
⇒ a₃ = 3x² - 3x + 3
a₃ = 3(1)² - 3(1) + 3
a₃ = 3(1) - 3 + 3
a₃ = 3
the difference of a term and the preceding term is same
i.e., 3 - 3 = 0
So, they're in A.P.
- Put x = 2,
⇒ a₁ = 2x + 1
a₁ = 2(2) + 1
a₁ = 4 + 1
a₁ = 5
⇒ a₂ = x² + x + 1
a₂ = 2² + 2 + 1
a₂ = 4 + 3
a₂ = 7
⇒ a₃ = 3x² - 3x + 3
a₃ = 3(2)² - 3(2) + 3
a₃ = 3(4) - 6 + 3
a₃ = 12 - 3
a₃ = 9
the difference of a term and the preceding term is same.
i.e., 9 - 7 = 7 - 5 = 2
So, they're in A.P.
Hence verified!