Further Mathematics Questions and Answers

Given an exponential sequence, say \(a, b, c,...\), as consecutive terms, then \(\sqrt{a \times c} = b\).

\(\therefore 2, (k+1), 8 \implies \sqrt{2 \times 8} = k + 1\)

\(k + 1 = \pm{4} \implies k = \text{-5 or 3}\)

To get the third factor, take the product of the other 2 factors and then divide the main equation by their product.

\((1.98)^{6} = (1 + 0.98)^{6} = 1 + 6(0.98) + 15(0.98)^{2} + 20(0.98)^{3} + 15(0.98)^{4} + 6(0.98)^{5} + (0.98)^{6}\)

 \(\approxeq 1 + 5.88 + 14.406 + 18.823 + 13.836 + 5.424 + 0.886 \)

= \(60.255\)

\(\frac{1 + \sqrt{8}}{3 - \sqrt{2}}\)

Rationalizing by multiplying through with \(3 + \sqrt{2}\),

\((\frac{1 + \sqrt{8}}{3 - \sqrt{2}})(\frac{3 + \sqrt{2}}{3 + \sqrt{2}}) = \frac{3 + \sqrt{2} + 3\sqrt{8} + 4}{9 - 2}\)

= \(\frac{3 + \sqrt{2} + 3\sqrt{4 \times 2} + 4}{7} \)

= \(\frac{7 + 7\sqrt{2}}{7} = 1 + \sqrt{2}\)

\(8^{x} ÷ (\frac{1}{4})^{y} = 1\)

\((2^{3})^{x} ÷ (2^{-2})^{y} = 2^{0}\)

\(2^{3x - (-2y)} = 2^{0}\)

\(\implies 3x + 2y = 0 .... (1)\)

\(\log_{2}(x - 2y) = 1\)

\( x - 2y = 2^{1} = 2 ..... (2)\)

Solving equations 1 and 2,

\(x = \frac{1}{2}, y = \frac{-3}{4}\)

\((x - y) = \frac{1}{2} - \frac{-3}{4} = \frac{5}{4}\)