Percubaan Matematik Tambahan SPM 2021 (Selangor), Kertas 1 (Soalan 5 & 6)

Soalan 5:
(a) Diberi suatu janjang geometri dengan sebutan-sebutan a, ar, ar², ar³, … , ar^n-2, ar^n-1 dan hasil tambah sebutan pertama ialah S_n.
Terbitkan rumus hasil tambah sebutan pertama, S_n bagi janjang geometri apabila |r| < 1. [3 markah]

(b) Seterusnya, cari nilai bagi n dengan keadaan hasil tambah 2n sebutan pertama adalah 127/128 kali hasil tambah n sebutan yang mempunyai nisbah sepunya -½. [4 markah]

Penyelesaian:
(a)

\begin{array}{l} S_{n} = a + a r + a r^{2} + a r^{3} + … + a r^{n - 2} + a r^{n - 1} … … … … (1) \\ (1) \times r : r S_{n} = a r + a r^{2} + a r^{3} + a r^{4} + … + a r^{n - 1} + a r^{n} … … … … (2) \\ (1) - (2), \\ S_{n} - r S_{n} = a - a r^{n} \\ S_{n} (1 - r) = a (1 - r^{n}) \\ S_{n} = \frac{a (1 - r^{n})}{(1 - r)}, | r | < 1 \end{array}

(b)

\begin{array}{l} S_{2 n} = \frac{127}{128} S_{n} \\ \frac{a (1 - {(- \frac{1}{2})}^{2 n})}{1 - (- \frac{1}{2})} = \frac{127}{128} (\frac{a (1 - {(- \frac{1}{2})}^{n})}{1 - (- \frac{1}{2})}) \\ \frac{a (1 - {(- \frac{1}{2})}^{2 n})}{\frac{3}{2}} = \frac{127}{128} (\frac{a (1 - {(- \frac{1}{2})}^{n})}{\frac{3}{2}}) \\ 128 (1 - {(- \frac{1}{2})}^{2 n}) = 127 (1 - {(- \frac{1}{2})}^{n}) \\ 128 - 128 {(- \frac{1}{2})}^{2 n} = 127 - 127 {(- \frac{1}{2})}^{n} \\ Let {(- \frac{1}{2})}^{n} = x \\ 128 - 128 x^{2} = 127 - 127 x \\ 128 x^{2} - 127 x - 1 = 0 \\ (128 x + 1) (x - 1) = 0 \\ x = - \frac{1}{128} or x = 1 \\ {(- \frac{1}{2})}^{n} = - \frac{1}{128} \\ {(- \frac{1}{2})}^{n} = {(- \frac{1}{2})}^{7} \\ n = 7 \\ atau {(- \frac{1}{2})}^{n} = 1 \\ {(- \frac{1}{2})}^{n} = {(- \frac{1}{2})}^{0} \\ n = 0 (tidak diterima) \end{array}

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