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Original question & working

Eliminate two progression parameters · source reference

[2024/NJC/Prelim/II/2] · Question pp. 2 · Working pp. 9, 10

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Printed question

An arithmetic series has first term aa and common difference dd, where a,da,d are non-zero. A convergent geometric series has first term bb and common ratio rr, where bb is positive and rr is non-zero. The fourth and ninth terms of the arithmetic series equal the sixth and ninth terms of the geometric series respectively, and the eleventh term of the arithmetic series is br6br^6 less than the fourteenth term of the geometric series.

(i) Show rr satisfies 5r8−7r3−5r+2=05r^8-7r^3-5r+2=0 and solve, giving your answer correct to 4 decimal places. [4]

(ii) Using this value of rr, deduce that for any positive integer nn, the sum of the terms of the geometric series after, but not including, term nn is less than 35b\frac35b. [2]

Printed working

(i) From AP un=a+(n−1)du_n=a+(n-1)d and GP un=brn−1u_n=br^{n-1},

a+3d=br5(1),a+8d=br8(2),a+10d=br13−br6(3).a+3d=br^5\quad(1),\qquad a+8d=br^8\quad(2),\qquad a+10d=br^{13}-br^6\quad(3).

(2)−(1)(2)-(1): d=b5(r8−r5)d=\frac b5(r^8-r^5). (3)−(2)(3)-(2): d=b2(r13−r8−r6)d=\frac b2(r^{13}-r^8-r^6).

b5(r8−r5)=b2(r13−r8−r6).\frac b5(r^8-r^5)=\frac b2(r^{13}-r^8-r^6).

Since b≠0,r≠0b\ne0,r\ne0,

2r8−2r5=5r13−5r8−5r6⇒2r3−2=5r8−5r3−5r⇒5r8−7r3−5r+2=0.2r^8-2r^5=5r^{13}-5r^8-5r^6\Rightarrow2r^3-2=5r^8-5r^3-5r\Rightarrow5r^8-7r^3-5r+2=0.

GC real roots: 1.1371 and 0.34347. Since convergent, r=0.3435r=0.3435 (4 d.p.).

(ii) Required sum =un+11−r=b(0.3435)n+1−11−0.3435≈1.523b(0.3435)n≤1.523b(0.3435)1≈0.52315b<35b=\frac{u_{n+1}}{1-r}=\frac{b(0.3435)^{n+1-1}}{1-0.3435}\approx1.523b(0.3435)^n\le1.523b(0.3435)^1\approx0.52315b<\frac35b.

Alternatively required sum =S∞−Sn=b1−r−b(1−rn)1−r=brn1−r=S_{\infty}-S_n=\frac b{1-r}-\frac{b(1-r^n)}{1-r}=\frac{br^n}{1-r}.

Printed answer key

(i) r=0.3435r=0.3435 (4 d.p.)

Mathematical review and qualifications

(i) Subtracting yields 5d=b(r8−r5)5d=b(r^8-r^5) and 2d=b(r13−r6−r8)2d=b(r^{13}-r^6-r^8). Equating 10d10d and cancelling br5≠0br^5 \ne0 gives 2(r3−1)=5(r8−r−r3)2(r^3-1)=5(r^8-r-r^3), hence 5r8−7r3−5r+2=05r^8-7r^3-5r+2=0. Its real roots are approximately 0.34347,1.13710.34347,1.1371; convergence rejects the latter, so r=0.3435r=0.3435 (4 d.p.).

(ii) Retain the unrounded root. The tail is brn/(1−r)≤br/(1−r)≈0.5232b<0.6b=3b/5br^n/(1-r)\le br/(1-r)\approx0.5232b<0.6b=3b/5.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2024/NJC/Prelim/II/2] Question PDF pages 2; solution PDF pages 9, 10. The original files remain private QA records outside this native content.