Arithmetic & Geometric Progressions: AP terms form GP; convergence; AP sigma proof — ASRJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
- The ninth, fifth and second terms of an arithmetic progression are successive terms of a geometric progression with first term \(a\) and common ratio \(r\), where \(r\neq 1\) and \(a>0\). Find the value of \(r\) and deduce that the geometric series is convergent.
- Using the value of \(r\) found in (i), find the least value of \(n\) such that the sum of all the terms after the \(n\)th term of the geometric progression is less than \(1\%\) of its sum of the first \(n\) terms.
- The sum, \(S_n\), of the first \(n\) terms of another sequence is given by \(S_n = n\ln 2^{q(n+1)}\), where \(q\) is a constant. Prove that the sequence follows an arithmetic progression.
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(a)(i) Find \(r\) and show the GP converges.
Let the first term and common difference of the AP be \(b\) and \(d\) respectively. The 9th, 5th and 2nd terms of the AP are \(b+8d\), \(b+4d\) and \(b+d\) respectively.
Since these are consecutive terms of a GP: \[ \frac{b+4d}{b+8d} = \frac{b+d}{b+4d} \] \[ (b+4d)^2 = (b+8d)(b+d) \] \[ b^2 + 8bd + 16d^2 = b^2 + 9bd + 8d^2 \] \[ d(8d - b) = 0 \]
Since \(d \neq 0\): \(b = 8d\). Therefore: \[ r = \frac{b+4d}{b+8d} = \frac{8d+4d}{8d+8d} = \frac{12d}{16d} = \frac{3}{4} \]
Since \(|r| = \left|\dfrac{3}{4}\right| < 1\), the geometric series is convergent.
(a)(ii) Least \(n\) such that sum of terms after \(n\)th term \(< 1\%\) of sum of first \(n\) terms.
The condition is \(S_\infty - S_n < 0.01 S_n\), i.e. \(S_\infty < 1.01 S_n\): \[\begin{aligned} \frac{a}{1-r} &< 1.01 \cdot \frac{a(1-(3/4)^n)}{1-3/4}\\ 1 &< 1.01\!\left(1 - \left(\frac{3}{4}\right)^n\right)\\ \left(\frac{3}{4}\right)^n &< \frac{0.01}{1.01} \approx 0.009901\\ n\ln\!\left(\frac{3}{4}\right) &< \ln(0.009901)\\ n &> \frac{\ln(0.009901)}{\ln(3/4)} \approx 16.042 \end{aligned}\]
(b) Prove the sequence with \(S_n = n\ln 2^{q(n+1)}\) is an AP.
The \(n\)th term: \[\begin{aligned} T_n &= S_n - S_{n-1}\\ &= n\ln 2^{q(n+1)} - (n-1)\ln 2^{q\cdot n}\\ &= n(n+1)q\ln 2 - (n-1)(nq)\ln 2\\ &= nq\ln 2\bigl[(n+1) - (n-1)\bigr]\\ &= 2nq\ln 2 \end{aligned}\]
The common difference: \[ T_n - T_{n-1} = 2nq\ln 2 - 2(n-1)q\ln 2 = 2q\ln 2 \]
This is constant (since \(q\) is a constant), so the sequence is an arithmetic progression. (Proved)