N2022 P1 Q9

N2022 P1 Q9

Junior College 2
8 marks
Free
  1. An arithmetic series has first term \(a\) and common difference \(d\), where \(d\ne0\). The first, third and fifteenth terms of this series are the first, second and third terms of a geometric series. Find \(d\) in terms of \(a\).[3]
  2. A geometric series has first term \(\sin\theta\) and common ratio \(-\cos\theta\), where \(0<\theta<\frac\pi2\).
    1. Show that the sum to infinity of this series is \(\tan k\theta\), where \(k\) is a constant to be found.[3]
    2. Given that \(\theta=\frac\pi3\), find the exact sum of the first seven terms of this series.[2]

Default solution

  1. \((a+2d)^2=a(a+14d)\) (consecutive geometric terms)
    \(a^2+4ad+4d^2=a^2+14ad\Rightarrow 2d(2d-5a)=0\)
    \(d\ne0\Rightarrow d=\frac52a\)
    1. \(S_\infty=\frac{\sin\theta}{1+\cos\theta}=\frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)}\)
      \(\therefore\ S_\infty=\tan\frac\theta2\Rightarrow k=\frac12\) (shown)
    2. \(\theta=\frac\pi3\Rightarrow a=\frac{\sqrt3}2,\quad r=-\frac12\)
      \(S_7=\frac{a(1-r^7)}{1-r}=\frac{\sqrt3}{2}\frac{1-(-1/2)^7}{1+1/2}=\frac{43\sqrt3}{128}\)
Answer:(a) \(d=\frac52a\) (b)(i) \(k=\frac12\) (b)(ii) \(\frac{43\sqrt3}{128}\)

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