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\(\frac{r}{a}\) + \(\frac{r}{b}\) + \(\frac{r}{c}\) > 1 = \(\frac{bcr + acr + abr}{abc}\) > 1 r(bc + ac + ba > abc) = r > \(\frac{abc}{bc + ac + ab}\)

-8, m, n, 19 = m + 8 = 19 - n m + n = 11 i.e. 1, 10

(a) \(T_{n} = a + (n - 1) d\) (terms of an A.P) Given: \(T_{8} - T_{4} = 20\) \(T_{8} = 1\frac{1}{2} \times T_{4}\) \(\therefore a + 7d - (a + 3d) = 20\) \(4d = 20 \implies d = 5\) (ii) Put d = 5 in the equation \(a + 7d = 1\frac{1}{2} \times (a + 3d)\) \(a + 7d = 1.5a + 4.5d\) \(7d - 4.5d = 1.5a - a \implies 2.5d = 0.5a\) \(a = \frac{2.5 \times 5}{0.5} = 25\) (b) \(A = P(1 - \frac{r}{100})^{n}\) where P = N10,250.00 r = 5% n = 9 \(\therefore A = (10,250) (1 - \frac{5}{100})^{9}\) = \(10,250 \times (0.95)^{9} = 10,250 \times 0.6302\) = \(N6,460.056 \approxeq N6,460\) (to 3 sig. figs)

(a) Roots of the equation = \(\frac{1}{2}\) and -7 \(\therefore (x - \frac{1}{2})(x + 7) = 0\) \(x^{2} - \frac{1}{2} x + 7x - \frac{7}{2} = 0\) \(x^{2} + \frac{13}{2} x - \frac{7}{2} = 0\) Multiply through by 2 to get integers as coefficients, \(2x^{2} + 13x - 7 = 0\) (b) Let the father and daughter's present ages be r and t respectively. Three years ago, father = (r - 3) years \((r - 3) = 4t \implies r = (4t + 3) ... (1)\) \(r \times t = 430 ... (2)\) Substitute (1) in (2), \((4t + 3) \times t = 430 \implies 4t^{2} + 3t = 430\) \(\implies 4t^{2} + 3t - 430 = 0\) \(4t^{2} - 40t + 43t - 430 = 0\) \(4t(t - 10) + 43(t - 10) = 0\) \(\therefore (4t + 43)(t - 10) = 0\) \(t = \frac{-43}{4}; t = 10\) Since t is age, it cannot be negative. Hence, t = 10 years. Recall : \(r = 4t + 3\) \(\therefore r = 4(10) + 3 = 40 + 3 = 43\) Therefore, the father's present age is 43 years while the daughter's age is 10 years.

Un = a + (n-1)d      = -3 + (8 - 1)2      = 11

The linear equation will be the difference between the equation. y=x2 + 4x - 6 and x2 + 4x - 7 = 0 y=x2 + 4x - 6, ( 0 = x2 + 4x - 7) y = -(-1) y = 1

x + 2 and x - 7 are factors (x+2)(x+7) = 0 x(x-7) + 2(x-7) = 0 x2 - 7x + 2x - 14 = 0 x2 - 5x - 14 = 0

In GP, when you are given three consecutive terms, say f, g, h, then \(f \times h = g^2\) Given: \(x, \frac{3}{2}, \frac{6}{7}, y\), then \(\frac{6x}{7} = (\frac{3}{2})^2 \implies \frac{6x}{7} = \frac{9}{4} ... (i)\) Also, \(\frac{3y}{2} = (\frac{6}{7})^2 \implies \frac{3y}{2} = \frac{36}{49} ... (ii)\) From \(\frac{6x}{7} = \frac{9}{4} \implies x = \frac{9 \times 7}{6 \times 4}\) \(x = \frac{21}{8}\) Also, \(\frac{3y}{2} = \frac{36}{49} \implies y = \frac{2 \times 36}{3 \times 49}\) = \(\frac{24}{49}\) \(xy = \frac{21}{8} \times \frac{24}{49} = \frac{9}{7}\)

\(x^{2} - x - 1 = 0\) Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) a = 1, b = -1, c = -1. \(x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(1)(-1)}}{2(1)}\) \(x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\) \(x = \frac{1 + 2.24}{2} ; x = \frac{1 - 2.24}{2}\) \(x = \frac{3.24}{2}; x = \frac{-1.24}{2}\) \(x = 1.62 ; x = -0.61 \) \(x \approxeq 1.6; -0.6\)

Let the first term and common difference = a & d respectively. \(T_{n} = \text{nth term} = a + (n - 1) d\) (A.P) Given: \(T_{4} = -6 \implies a + 3d = -6 ... (i)\) \(T_{8} + T_{9} = 72\) \(\implies a + 7d + a + 8d = 72 \implies 2a + 15d = 72 ... (ii)\) From (i), \(a = -6 - 3d\) \(\therefore\) (ii) becomes \(2(-6 - 3d) + 15d = 72\) \(-12 - 6d + 15d = 72 \implies 9d = 72 + 12 = 84\) \(d = \frac{84}{9} = 9\frac{1}{3}\)

\(T_{n} = ar^{n - 1}\) (nth term of a G.P) \(T_{4} = ar^{3} = 189\) \(7 \times r^{3} = 189 \implies r^{3} = 27\) \(r = \sqrt[3]{27} = 3\) \(S_{n} = \frac{a(r^{n} - 1)}{r - 1}\) \(S_{3} = \frac{7(3^{3} - 1)}{3 - 1} \) = \(\frac{7 \times 26}{2} = 91\)

(m + \(\sqrt{3}\))(1 - \(\sqrt{3}\))2 = 6 - 2\(\sqrt{2}\) (m + \(\sqrt{3}\))(4 - 2\(\sqrt{3}\)) = 6 - 2\(\sqrt{2}\) = 6 - 2\(\sqrt{3}\) 4m - 6 + 4 - 2m\(\sqrt{3}\) = 6 - 2\(\sqrt{3}\) comparing co-efficients, 4m - 6 = 6.......(i) 4 - 2m = -2.......(ii) in both equations, m = 3

\(x^2 + \frac{2x}{r^2}\) + \(\frac{1}{r^4}\) = 0 (x + \(\frac{1}{r^2}\)) = 0 x + \(\frac{1}{r^2}\) = 0 x = \(\frac{-1}{r^2}\)

-1 < 5 - 2x \(\geq\) 7 = -1 < 5 -2x and 5 - 2x \(\leq\) 7 = 2x < 5 + 1 and 5 - 7 \(\leq\) 2x = x < 3 and -1 \(\leq\) x Integral value of x are -1, 0, 1, 2

3x - 5y = 3, 2y - 6x = -5 -5y + 3x = 3........{i} x 2 2y - 6x = -5.........{ii} x 5 Substituting for x in equation (i) -5y + 3(\(\frac{19}{24}\)) = 3 -5y + 3 x \(\frac{19}{24}\) = 3 -5y = \(\frac{3 - 19}{8}\) -5 = \(\frac{24 - 19}{8}\) = \(\frac{5}{8}\) y = \(\frac{5}{8 \times 5}\) y = \(\frac{-1}{8}\) (x, y) = (\(\frac{19}{24}, \frac{-1}{8}\)