Question

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​

Answer 1

EXPLANATION.

(1) = 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).

Option [C] is correct answer.

(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.

Option [B] is correct answer.

Answer 2

$\large\underline{\text{Solution, Question 1}}$

By Vieta's formulas

$\implies\alpha+\beta=3$

$\implies\alpha\beta=a$

$\implies\gamma+\delta=12$

$\implies\gamma\delta=b$

Consider

$\implies\alpha,\beta,\gamma,\delta\text{ (G.P)}$

$\implies k,kr,kr^2,kr^3\text{ (G.P)}$

The sum of two consecutive terms are

$\implies k(r+1)=3$

$\implies kr^2(r+1)=12$

On dividing while the common ratio is positive

$\implies r^2=4$

$\implies r=2\ (\because r>0)$

Furthermore

$\implies k=1$

Since $\alpha,\beta,\gamma,\delta$ form an increasing G.P

$\implies\alpha=1,\beta=2,\gamma=4,\delta=8$

Hence

$\implies a=2,b=32$

$\large\underline{\text{Answer, Question 1}}$

The correct option is choice C.

$\large\underline{\text{Solution, Question 2}}$

Here we consider two sequences.

$\implies S_{1}=1^{2}+2^{2}+3^{2}+\cdots+28^{2}+29^{2}+30^{2}$

$\implies S_{2}=2\cdot(3^{2}+6^{2}+9^{2}+\cdots+24^{2}+27^{2}+30^{2})$

Then, $S_{1}-S_{2}$ is the required answer.

Factorization of $S_{2}$ is

$\implies S_{2}=2\cdot3^{2}\cdot(1^{2}+2^{2}+3^{2}+\cdots+8^{2}+9^{2}+10^{2})$

$\red{\bigstar}$The sum of the perfect squares.

$\implies\displaystyle\sum^{n}_{k=1}k^{2}=\dfrac{n(n+1)(2n+1)}{6}$

Now we have

$\implies S_{1}=\dfrac{30\cdot31\cdot61}{6}=9455$

$\implies S_{2}=18\cdot\dfrac{10\cdot11\cdot21}{6}=6930$

Thus

$\implies S_{1}-S_{2}=9455-6930=2525$

$\large\underline{\text{Answer, Question 2}}$

The correct option is choice B.