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Concept 04 / 07Sequences and Series

Geometric Progressions: nth Term, Finite and Infinite Sums

The nth term ar^(n-1), the sum a(r^n - 1)/(r - 1) by the shift-and-subtract method, the infinite sum a/(1 - r) and its condition |r| < 1, recurring decimals, properties of GPs and the geometric mean.

Figure 1
A geometric progression (GP) has a constant ratio between consecutive terms: Tn+1Tn=r\frac{T_{n+1}}{T_{n}} = r, with a≠0a \neq 0 and r≠0r \neq 0. Its general term is Tn=arn−1T_{n} = ar^{n-1}. Where an AP grows linearly, a GP grows (or decays) exponentially, which is why GPs model compound interest, population growth and radioactive decay. Sum of nn terms: the shift-and-subtract trick. Let S=a+ar+⋯+arn−1S = a + ar + \cdots + ar^{n-1}. Then rS=ar+ar2+⋯+arnrS = ar + ar^{2} + \cdots + ar^{n}. Subtracting, every middle term cancels: S−rS=a−arnS - rS = a - ar^{n}. So, for r≠1r \neq 1, Sn=a(1−rn)1−r=a(rn−1)r−1S_{n} = \frac{a(1 - r^{n})}{1 - r} = \frac{a(r^{n} - 1)}{r - 1}. Use the first form when ∣r∣<1|r| < 1 and the second when ∣r∣>1|r| > 1 to keep signs positive. If r=1r = 1, every term is aa and Sn=naS_{n} = na; the formula is undefined, which is a frequent trap in parameter questions. Infinite GP. If ∣r∣<1|r| < 1, then rn→0r^{n} \to 0 as n→∞n \to \infty, so S∞=a1−rS_{\infty} = \frac{a}{1 - r}. If ∣r∣≥1|r| \ge 1 (and a≠0a \neq 0), the terms do not shrink to zero and the series has no finite sum. The figure shows 12+14+18+⋯\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots filling a unit bar: each piece covers half of what remains, so the uncovered part 12n\frac{1}{2^{n}} tends to 00 and the sum is 11. For r=−1r = -1 the partial sums oscillate between aa and 00; for ∣r∣>1|r| > 1 they grow without bound. Recurring decimals are infinite GPs. 0.7‾=710+7100+⋯=7/101−1/10=790.\overline{7} = \frac{7}{10} + \frac{7}{100} + \cdots = \frac{7/10}{1 - 1/10} = \frac{7}{9}. In general, a block of kk repeating digits BB after the decimal point is B10k−1\frac{B}{10^{k} - 1}, and a non-repeating part is handled by shifting: 0.123‾=110+110⋅2399=122990=614950.1\overline{23} = \frac{1}{10} + \frac{1}{10} \cdot \frac{23}{99} = \frac{122}{990} = \frac{61}{495}. Properties. Multiplying or dividing every term by a non-zero constant gives a GP with the same ratio. Reciprocals of a GP form a GP with ratio 1r\frac{1}{r}; powers TnkT_{n}^{k} form a GP with ratio rkr^{k}. Term-by-term products and quotients of GPs are GPs; sums generally are not. a,b,ca, b, c (non-zero) are in GP if and only if b2=acb^{2} = ac. The product of terms equidistant from both ends is constant: TkTn−k+1=T1TnT_{k}T_{n-k+1} = T_{1}T_{n}. If the terms are positive, their logarithms form an AP: log⁡Tn=log⁡a+(n−1)log⁡r\log T_{n} = \log a + (n - 1)\log r. This connection lets AP results transfer to GPs. Choosing terms symmetrically: three terms ar,a,ar\frac{a}{r}, a, ar (product a3a^{3}); four terms ar3,ar,ar,ar3\frac{a}{r^{3}}, \frac{a}{r}, ar, ar^{3} (ratio r2r^{2}, product a4a^{4}). Geometric mean. For positive aa and bb, G=abG = \sqrt{ab} makes a,G,ba, G, b a GP. Inserting nn geometric means between aa and bb gives ratio r=(ba)1/(n+1)r = (\frac{b}{a})^{1/(n+1)}, and the product of the nn means is (ab)n(\sqrt{ab})^{n}. Key Takeaways: - Tn=arn−1T_{n} = ar^{n-1}; Sn=a(rn−1)r−1S_{n} = \frac{a(r^{n} - 1)}{r - 1} for r≠1r \neq 1; Sn=naS_{n} = na for r=1r = 1. - S∞=a1−rS_{\infty} = \frac{a}{1 - r} only for ∣r∣<1|r| < 1. - b2=acb^{2} = ac for a,b,ca, b, c in GP; TkTn−k+1=T1TnT_{k}T_{n-k+1} = T_{1}T_{n}. - Logs of a positive GP form an AP. - Symmetric choices: ar,a,ar\frac{a}{r}, a, ar. - Product of nn geometric means between aa and bb: (ab)n/2(ab)^{n/2}. - Recurring decimal with block BB of length kk: B10k−1\frac{B}{10^{k} - 1}. - Pitfall: applying a1−r\frac{a}{1 - r} when ∣r∣≥1|r| \ge 1, or forgetting the case r=1r = 1. Brain-Teaser 1: The sum of an infinite GP is 1515 and the sum of the squares of its terms is 4545. Find the GP. Solution: a1−r=15\frac{a}{1 - r} = 15 and, since the squares form a GP with first term a2a^{2} and ratio r2r^{2}, a21−r2=45\frac{a^{2}}{1 - r^{2}} = 45. Divide the second by the square of the first: a2/(1−r2)a2/(1−r)2=1−r1+r=45225=15\frac{a^{2}/(1 - r^{2})}{a^{2}/(1 - r)^{2}} = \frac{1 - r}{1 + r} = \frac{45}{225} = \frac{1}{5}. So 5−5r=1+r5 - 5r = 1 + r, giving r=23r = \frac{2}{3} and a=15×13=5a = 15 \times \frac{1}{3} = 5. The GP is 5,103,209,…5, \frac{10}{3}, \frac{20}{9}, \ldots. Dividing the two equations eliminates aa in one step. Brain-Teaser 2: Find the sum of nn terms of 0.7+0.77+0.777+⋯0.7 + 0.77 + 0.777 + \cdots. Solution: The kkth term is 79(1−10−k)\frac{7}{9}(1 - 10^{-k}), since 0.77…70.77\ldots7 with kk sevens is 79×0.99…9\frac{7}{9} \times 0.99\ldots9. So the sum is 79[n−∑k=1n10−k]=79[n−19(1−10−n)]=781(9n−1+10−n)\frac{7}{9}\left[n - \sum_{k=1}^{n}10^{-k}\right] = \frac{7}{9}\left[n - \frac{1}{9}(1 - 10^{-n})\right] = \frac{7}{81}(9n - 1 + 10^{-n}). Rewriting each term as a constant minus a GP term is the standard technique for such sequences. Brain-Teaser 3: If a,b,ca, b, c are in AP and also in GP, show that a=b=ca = b = c. Solution: From the AP, 2b=a+c2b = a + c; from the GP, b2=acb^{2} = ac. Then (a−c)2=(a+c)2−4ac=4b2−4b2=0(a - c)^{2} = (a + c)^{2} - 4ac = 4b^{2} - 4b^{2} = 0, so a=ca = c, and then b=a+c2=ab = \frac{a + c}{2} = a. So the only sequences that are both an AP and a GP are constant (with non-zero terms). This is often used silently in multiple-choice questions.

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Geometric Progressions: nth Term, Finite and Infinite Sums — DPP

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