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Original question & working

Selected GP terms form an AP · source reference

TJC JC2 2025 H2 Maths MSM; no separate exam attribution printed for Q5 · Question pp. 1 · Working pp. 6, 7

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The printed solution divides by r^4 and thereby omits the zero-ratio GP. The displayed main answer follows its intended nondegenerate branch.

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Printed question

The fifth, tenth and twentieth terms of a convergent geometric progression, G, are the first three consecutive terms of an arithmetic progression, A.

(i) Determine the common ratio of G. [3]

Given that the first term of G is 2,

(ii) evaluate the sum to infinity for G [2];

(iii) find the sum of the first 10 odd-numbered terms in A. [3]

Printed working

(i) Let aa be the first term of G.

ar9−ar4=ar19−ar9⇒ar4(r5−1)=ar9(r10−1),ar^9-ar^4=ar^{19}-ar^9\Rightarrow ar^4(r^5-1)=ar^9(r^{10}-1), r5−1(r5)2−1=r5⇒r5−1(r5−1)(r5+1)=r5⇒1r5+1=r5⇒r10+r5−1=0.\frac{r^5-1}{(r^5)^2-1}=r^5\Rightarrow\frac{r^5-1}{(r^5-1)(r^5+1)}=r^5\Rightarrow\frac1{r^5+1}=r^5\Rightarrow r^{10}+r^5-1=0.

From GC: r=−1.10r=-1.10 (rejected since convergence requires ∣r∣<1|r|<1) or r=0.90824r=0.90824 (5 s.f.), hence 0.908 (3 s.f.).

(ii) S∞=21−0.90824=21.8S_{\infty}=\frac2{1-0.90824}=21.8 (3 s.f.).

(iii) AP first term =2r4=2r^4, common difference =2r9−2r4=2r^9-2r^4. The odd-term sum is

102[2(2r4)+9⋅2(2r9−2r4)]=−33.2.\frac{10}2[2(2r^4)+9\cdot2(2r^9-2r^4)]=-33.2.

Teacher notes

For GP G, Gn=arn−1G_n=ar^{n-1}. AP terms A1,A2,A3A_1,A_2,A_3 are G5,G10,G20=ar4,ar9,ar19G_5,G_{10},G_{20}=ar^4,ar^9,ar^{19}. Equal differences: A2−A1=A3−A2A_2-A_1=A_3-A_2.

AP odd-indexed terms A1,A3,A5,…A_1,A_3,A_5,\ldots have difference 2d2d. “Odd-numbered” refers to indices: their values need not be odd.

GP infinite sum a/(1−r)a/(1-r).

Printed answer key

(i) r=0.90824r=0.90824 (5 s.f.) =0.908 (3 s.f.); (ii) 21.8 (3 s.f.); (iii) −33.2

Mathematical review and qualifications

For first term bb, equal differences give br4(r15−2r5+1)=0br^4(r^{15}-2r^5+1)=0. In the intended nondegenerate case b≠0,r≠0b \ne0,r \ne0, put t=r5t=r^5: t3−2t+1=(t−1)(t2+t−1)=0t^3-2t+1=(t-1)(t^2+t-1)=0. Convergence excludes r=1r=1 and the negative root with modulus exceeding one. Thus r=[(5−1)/2]1/5≈0.90824r=[(\sqrt5-1)/2]^{1/5}\approx0.90824.

(ii) With b=2b=2, S∞=2/(1−r)≈21.8S_\infty=2/(1-r)\approx21.8 (3 s.f.).

(iii) A1=2r4A_1=2r^4, d=2r9−2r4d=2r^9-2r^4. The ten odd-indexed terms have difference 2d2d, sum 10[A1+9d]≈−33.210[A_1+9d]\approx-33.2.

Qualification: the literal wording also admits r=0r=0, yielding G=(2,0,0,…)G=(2,0,0,\ldots) and the zero AP. Then (ii) is 2 and (iii) is 0. The printed solution chooses the nonzero branch; that assumption is recorded rather than silently imposed.

The printed solution divides by r^4 and thereby omits the zero-ratio GP. The displayed main answer follows its intended nondegenerate branch.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. TJC JC2 2025 H2 Maths MSM; no separate exam attribution printed for Q5 Question PDF pages 1; solution PDF pages 6, 7. The original files remain private QA records outside this native content.