Further Mathematics Questions and Answers

\(S_{n} = \frac{n}{2}(2a + (n - 1) d = n^{2} + 2n\)

\(n(2a + (n - 1) d = 2n^{2} + 4n\)

\(2an + n^{2}d - nd = 2n^{2} + 4n\)

\(n^{2}d = 2n^{2}\)

\(d = 2\)

\((2a - d) n = 4n\)

\(2a - d = 4 \implies 2a = 4 + d = 4 + 2 = 6\)

\(a = 3\)

\(T_{n} = a + (n - 1)d\)

= \(3 + (n - 1)2 = 3 + 2n - 2 = 2n + 1\)

\(S_{n} = \frac{n}{2}(2a + (n - 1)d \)

\(\frac{n}{2}(2a + (n - 1) d = n^{2} + 2n\)

\(n(2a + (n - 1) d = 2n^{2} + 4n\)

\(2an + n^{2}d - nd = 2n^{2} + 4n\)

Comparing the equations, d = 2 (coefficient of \(n^{2}\)).

\(f(x) = \frac{x + 3}{x - 2}\) 

\(f(y) = \frac{y + 3}{y - 2}\)

Let f(y) = x,

\(x = \frac{y + 3}{y - 2}\)

\(x(y - 2) = y + 3\)

\(xy - y = 2x + 3 \implies y(x - 1) = 2x + 3\)

\(y = \frac{2x + 3}{x - 1}\)

\(2x^{2} + 7x - 15 > 0 \implies 2x^{2} - 3x + 10x - 15 > 0\)

\(x(2x - 3) + 5(2x - 3) > 0\)

\((x + 5)(2x - 3) > 0\)

For their product to be positive, they are either both +ve or -ve.

\(x + 5 > 0 \implies x > -5\)

\(2x - 3 > 0 \implies 2x > 3\)

\(x > \frac{3}{2}\)

Check:

\(x > -5: x = -3\)

\(2(-3)^{2} + 7(-3) - 15 = 18 - 21 - 15 = -18 < 0\) (Not satisfied)

\(\therefore x < -5\)

\(x > \frac{3}{2}: x = 2\)

\(2(2^{2}) + 7(2) - 15 = 8 + 14 - 15 = 7 > 0\) (Satisfied)

\(\frac{\sqrt{3} + \sqrt{48}}{\sqrt{6}} = \frac{\sqrt{3} + 4\sqrt{3}}{\sqrt{6}}\)

\(\frac{5\sqrt{3}}{\sqrt{6} = \frac{5\sqrt{3} \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}}\)

\(\frac{5\sqrt{18}{6} = \frac{15\sqrt{2}}{6}\)

= \(\frac{5\sqrt{2}}{2}\)