(a) (i) ar\(^2\) = \(\frac{1}{4}\) ........(i) ar\(^5\) = \(\frac{1}{32}\) ........(ii) \(\frac{ar^5}{ar^2} = \frac{1}{32} \div \frac{1}{4}\) r\(^3 = \frac{1}{32} \times \frac{4}{1}\) r\(^3 = \frac{1}{32} \times \frac{4}{1}\) r\(^3 = 2^{-3} = (\frac{1}{2})^3\) r = \(\frac{1}{2}\) Since ar\(^2\) = \(\frac{1}{4}\) a(\(\frac{1}{2}\))\(^2\) = \(\frac{1}{4}\) a = 1 The first term is 1 ad the common ratio is \(\frac{1}{2}\) (ii) T\(_7\) = ar\(^6\) = 1 x (\(\frac{1}{2}\))\(^6\) The seventh term = \(\frac{1}{64}\) (b) x = 2 and x = -3 a(2)\(^2\) + b(2) + c = 0 4a + 2b + c = 0 4a + 2b + c = 0.....(i) a(-3)\(^2\) + b(-3) + c = 0 9a - 3b + c = 0.......(ii) Solving eq.(i) and (ii) 4a + 2b + c = 0 -(9a - 3b + c) = 0 -5a + 5b = 0 (x - 2)(x + 3) = 0 x\(^2\) + 3x - 2x - 6 = 0 x\(^2\) + x - 6 = 0 a = 1, b = 1, c = -6