1) Given that α , β are the roots of the equation x² - 3x + a = 0 and γ , δ are the roots of the equation x² - 12x + b= 0. If α , β , γ , δ form an increasing G. P., then the ordered pair (a, b) is equal to:
A) (3 , 12)
B) (12 , 3)
C) (2 , 32)
D) (4 , 16)
2) Find the sum of the series upto 30 terms :
1² + 2² - 3² + 4² + 5² - 6² + 7² + 8² - 9² + ....
A) 2425 B) 2525 C) 2524 D) 2562
Answers
Answer:
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1) Given that α , β are the roots of the equation x² - 3x + a = 0 and γ , δ are the roots of the equation x² - 12x + b= 0. If α , β , γ , δ form an increasing G. P., then the ordered pair (a, b) is equal to:
- A) (3 , 12)
- B) (12 , 3)
- C) (2 , 32)
- D) (4 , 16)
Solution :-
α, β are the roots of the equation.
⇒ x² - 3x + a = 0.
γ, δ are the roots of the equation.
⇒ x² - 12x + b = 0.
As we know that,
We can write equation as,
⇒ α = a.
⇒ β = ar.
⇒ γ = ar².
⇒ δ = ar³.
α, β are the roots of the equation.
⇒ x² - 3x + a = 0.
Sum of the zeroes of the quadratic polynomial.
⇒ α + β = - b/a.
⇒ α + β = - (-3/1) = 3.
⇒ α + β = 3.
⇒ a + ar = 3.
⇒ a(1 + r) = 3. - - - - - (1).
Products of the zeroes of the quadratic polynomial.
⇒ αβ = c/a.
⇒ (a)(ar) = A. - - - - - (2).
γ, δ are the roots of the equation.
⇒ x² - 12x + b = 0.
As we know that,
Sum of the zeroes of the quadratic polynomial.
⇒ γ + δ = - b/a.
⇒ γ + δ = - (-12/1) = 12.
⇒ γ + δ = 12.
⇒ ar² + ar³ = 12.
⇒ ar²(1 + r) = 12. - - - - - (3).
Products of the zeroes of the quadratic polynomial.
⇒ γδ = c/a.
⇒ γδ = b.
⇒ (ar²)(ar³) = b. - - - - - (4).
From equation (1) and (3), we get.
Divide equation (1) and (3), we get.
⇒ a(1 + r) = 3. - - - - - (1).
⇒ ar²(1 + r) = 12. - - - - - (3).
We get.
⇒ r² = 4.
⇒ r = √4.
⇒ r = ± 2.
⇒ r = 2 [we included].
Put the value of r = 2 in the equation (1), we get.
⇒ a(1 + r) = 3.
⇒ a(1 + 2) = 3.
⇒ 3a = 3.
⇒ a = 1.
Put the values of a = 1 and r = 2 in the equation (2), we get.
⇒ (a)(ar) = A.
⇒ (1)(1 x 2) = A.
⇒ A = 2.
Put the values of a = 1 and r = 2 in the equation (2), we get.
⇒ (ar²)(ar³) = b.
⇒ [1 x (2)²] x [1 x (2)³] = b.
⇒ 4 x 8 = b.
⇒ b = 32.
Values of a = 2 and b = 32.
Ordered pair (a, b) = (2, 32).
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2) Find the sum of the series upto 30 terms : 1² + 2² - 3² + 4² + 5² - 6² + 7² + 8² - 9² + ....
- A) 2425
- B) 2525
- C) 2524
- D) 2562
Solution :-
(2) = 1² + 2² - 3² + 4² + 5² - 6² + 7² + 8² - 9² + . . . . .
As we know that,
We can write equation as,
⇒ (1² + 4² + 7² + . . . . . to 10 terms) + [(2² - 3³) + (5² - 6²) + (8² - 9²) + . . . . . to 20 term].
⇒ (1² + 4² + 7² + . . . . . to 10 terms) + [- 5 - 11 - 17 - . . . . . to 10 terms].
⇒ S₁ = (1² + 4² + 7² + . . . . . to 10 terms)
As we know that,
Formula of :
General term of an A.P.
⇒ Tₙ = a + (n - 1)d.
First term = a = 1.
Common difference = d = b - a = 4 - 1 = 3.
Using this formula in the equation, we get.
⇒ Tₙ = 1 + (n - 1)3.
⇒ Tₙ = 1 + 3n - 3.
⇒ Tₙ = 3n - 2.
⇒ Tₙ = (3n - 2)².
⇒ Tₙ = 9n² + 4 - 12n.
⇒ S₁ = 9 ∑n² - 12 ∑n + 4 ∑1.
As we know that,
Formula of :
⇒ ∑n² = [n(n + 1)(2n + 1)/6].
⇒ ∑n = [n(n + 1)/2].
⇒ ∑1 = n.
Using this formula in the equation, we get.
Put the value of n = 10 in the equation, we get.
⇒ S₁ = 9 x [n(n + 1)(2n + 1)/6] - 12 x [n(n + 1)/2] + 4 x n.
⇒ S₁ = 9 x [10(11)(21)/6] - 12(10(11)/2] + 4 x 10.
⇒ S₁ = 3465 - 660 + 40.
⇒ S₁ = 2845.
⇒ (1² + 4² + 7² + . . . . . to 10 terms) + [- 5 - 11 - 17 - . . . . . to 10 terms].
⇒ 2845 + [- 5 - 11 - 17 - . . . . . to 10 terms].
⇒ 2845 - [5 + 11 + 17 + . . . . . to 10 terms].
As we know that,
Formula of :
Sum of n terms of an A.P.
⇒ Sₙ = n/2[2a + (n - 1)d].
First term = a = 5.
Common difference = d = b - a = 11 - 5 = 6.
Put the values in the equation, we get.
⇒ S₁₀ = 10/2[2 x 5 + (10 - 1)6].
⇒ S₁₀ = 5[10 + 54].
⇒ S₁₀ = 5 x 64.
⇒ S₁₀ = 320.
⇒ (1² + 4² + 7² + . . . . . to 10 terms) + [- 5 - 11 - 17 - . . . . . to 10 terms].
⇒ 2845 - 320 = 2525.
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