Let a, b and c be three real numbers satisfying

Linked Question 1

Let b = 6, with a and c satisfying (E). If a and b are the roots of the quadratic equation ax² + bx + c= 0, then $\sum_{n = 0}^{\infty} {(\frac{1}{α} + \frac{1}{β})}^{n}$ , is

$\frac{6}{7}$

$\infty$

$∴ (a, b, c) \equiv (- \frac{λ}{7}, - \frac{6 λ}{7}, λ)$

$∴$ b = 6 so l = – 7.

So (a, b, c) $\equiv$ (1, 6, –7)

So the equation ax² + bx + c = 0

x² + 6x – 7 = 0

So $α$ = 1, B = – 7

$S = \sum_{n = 0}^{\infty} {(\frac{1}{α} + \frac{1}{β})}^{n} = \sum {(\frac{1}{1} - \frac{1}{7})}^{n}$

$= \sum {(\frac{6}{7})}^{n} = 1 + \frac{6}{7} + {(\frac{6}{7})}^{2} + . ..... \infty$

$= \frac{1}{1 - \frac{6}{7}} = 7$

Linked Question 2

Let ω be a solution of x³ – 1 = 0 with Im(ω) > 0. If a = 2 with b and c satisfying (E), then the value of $\frac{3}{ω^{a}} + \frac{1}{ω^{b}} + \frac{3}{ω^{c}}$ is equal to :

–2

–3

$∴ (a, b, c) \equiv (- \frac{λ}{7}, - \frac{6 λ}{7}, λ)$

$∴$ a = 2 is given so $λ$ = – 14

So (a, b, c) $\equiv$ (2, 12, – 14)

So, $\frac{3}{ω^{a}} + \frac{1}{ω^{b}} + \frac{3}{ω^{c}} = - 2$

Linked Question 3

If the point P(a, b, c), with reference to (E), lies on the plane 2x + y + z = 1, then the value of 7a + b + c, is

$[a b c] | \begin{matrix} 1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7 \end{matrix} | = [0 0 0]$

a + 8b + 7c = 0

9a + 2b + 3c = 0

7a + 7b + 7c = 0

On solving above equation

$(a, b, c) \equiv (- \frac{λ}{7}, - \frac{6 λ}{7}, λ)$

$∴$ (a, b, c) lies on the plane 2x + y + z = 1

So $- \frac{2 λ}{7} - \frac{6 λ}{7} + λ = 0$

on solving $λ$ = – 7

So 7a + b + c = 6