Predicting What Comes Next: Sequences and Progressions | EOT

Question 11

The sum of the first three terms of a GP is 1312\dfrac{13}{12} and their product is −1-1. Find the common ratio and the terms.

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Solution
Understand the Question
  • Let the three consecutive terms of a Geometric Progression (GP) be ar\dfrac{a}{r}, aa, and arar, where aa is the middle term and rr is the common ratio.
  • Using the product of the terms eliminates rr, allowing us to directly solve for aa.
  • Substituting aa into the sum of the terms gives a quadratic equation in rr, which yields two possible values for the common ratio and corresponding sets of terms.

Step 1 · Find the Middle Term

Let the three terms of the GP be ar\dfrac{a}{r}, aa, and arar.

Given their product is −1-1 (ar)×a×ar=a3=−1\left(\dfrac{a}{r}\right) \times a \times ar = a^3 = -1

a3=−1a=−13=−1\begin{aligned} a^3 &= -1 \\[0.6em] a &= \sqrt[3]{-1} = -1 \end{aligned}

Step 2 · Find the Common Ratio

Substituting a=−1a = -1, the terms become −1r-\dfrac{1}{r}, −1-1, and −r-r.

Given the sum of the terms is 1312\dfrac{13}{12} −1r−1−r=1312-\dfrac{1}{r} - 1 - r = \dfrac{13}{12}

Multiply the entire equation by 12r12r −12−12r−12r2=13r-12 - 12r - 12r^2 = 13r

12r2+12r+13r+12=012r2+25r+12=0\begin{aligned} 12r^2 + 12r + 13r + 12 &= 0 \\[0.6em] 12r^2 + 25r + 12 &= 0 \end{aligned}

Factorise by splitting the middle term

12r2+16r+9r+12=04r(3r+4)+3(3r+4)=0(4r+3)(3r+4)=0\begin{aligned} 12r^2 + 16r + 9r + 12 &= 0 \\[0.6em] 4r(3r + 4) + 3(3r + 4) &= 0 \\[0.6em] (4r + 3)(3r + 4) &= 0 \end{aligned} 4r+3=0or3r+4=0r=−34orr=−43\begin{aligned} 4r + 3 = 0 \quad &\text{or} \quad 3r + 4 = 0 \\[0.6em] r = -\dfrac{3}{4} \quad &\text{or} \quad r = -\dfrac{4}{3} \end{aligned}

Step 3 · Find the Terms of the GP

Case 1: When r=−34r = -\dfrac{3}{4} and a=−1a = -1

ar=−1−34=43a=−1ar=−1×(−34)=34\begin{aligned} \dfrac{a}{r} &= \dfrac{-1}{-\frac{3}{4}} = \dfrac{4}{3} \\[0.6em] a &= -1 \\[0.6em] ar &= -1 \times \left(-\dfrac{3}{4}\right) = \dfrac{3}{4} \end{aligned}

The terms are 43,−1,34\dfrac{4}{3}, -1, \dfrac{3}{4}.

Case 2: When r=−43r = -\dfrac{4}{3} and a=−1a = -1

ar=−1−43=34a=−1ar=−1×(−43)=43\begin{aligned} \dfrac{a}{r} &= \dfrac{-1}{-\frac{4}{3}} = \dfrac{3}{4} \\[0.6em] a &= -1 \\[0.6em] ar &= -1 \times \left(-\dfrac{4}{3}\right) = \dfrac{4}{3} \end{aligned}

The terms are 34,−1,43\dfrac{3}{4}, -1, \dfrac{4}{3}.

Answer

Common ratio r=−34r = -\dfrac{3}{4} or −43-\dfrac{4}{3}; the terms are 43,−1,34\dfrac{4}{3}, -1, \dfrac{3}{4} or 34,−1,43\dfrac{3}{4}, -1, \dfrac{4}{3}.

Common Mistakes
  • Assuming a,ar,ar2a, ar, ar^2 directly: While valid, assuming terms as a,ar,ar2a, ar, ar^2 leads to simultaneous non-linear equations which are harder to solve compared to using ar,a,ar\dfrac{a}{r}, a, ar.
  • Sign Errors in Multiplication: Forgetting that multiplying −1−1r−r=1312-1 - \dfrac{1}{r} - r = \dfrac{13}{12} by 12r12r introduces negative signs across all terms on the LHS.
  • Discarding One Solution: Neglecting either r=−34r = -\dfrac{3}{4} or r=−43r = -\dfrac{4}{3}. Both values of rr are valid real ratios that simply reverse the order of the GP.

More questions in EOT

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Q2

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Q3

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Q4

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Q5

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Q6

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Q7

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Q8

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Q9

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Q10

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Q11

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Q12

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Q13

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Q14

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Q15

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