Predicting What Comes Next: Sequences and Progressions | EOT

Question 11

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

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Solution

We will use the given product and sum to find the terms.

Step 1 — Find the middle term

Let the three terms of the GP be a/ra/r, aa, and arar. Let's find their product.

(a/r)×a×ar(a/r) \times a \times ar

=a3= a^3

The problem states the product is −1\mathbf{-1}.

a3=−1a^3 = -1

a=−13a = \sqrt[3]{-1}

a=−1\boxed{a = -1}

Step 2 — Find the common ratio

Now we know the middle term is −1\mathbf{-1}. The terms are −1/r-1/r, −1-1, and −r-r. Let's find their sum.

−1/r−1−r-1/r - 1 - r

The problem states the sum is 13/12\mathbf{13/12}.

−1/r−1−r=13/12-1/r - 1 - r = 13/12

Multiply all terms by 12r12r.

−12−12r−12r2=13r-12 - 12r - 12r^2 = 13r

Rearrange into a quadratic equation.

12r2+12r+13r+12=012r^2 + 12r + 13r + 12 = 0

12r2+25r+12=012r^2 + 25r + 12 = 0

Let's factor the quadratic equation.

12r2+16r+9r+12=012r^2 + 16r + 9r + 12 = 0

4r(3r+4)+3(3r+4)=04r(3r + 4) + 3(3r + 4) = 0

(4r+3)(3r+4)=0(4r + 3)(3r + 4) = 0

This gives two possible values for rr.

4r+3=0or3r+4=04r + 3 = 0 \quad \text{or} \quad 3r + 4 = 0

4r=−3or3r=−44r = -3 \quad \text{or} \quad 3r = -4

r=−3/4orr=−4/3r = -3/4 \quad \text{or} \quad r = -4/3

r=−3/4 or r=−4/3\boxed{r = -3/4 \text{ or } r = -4/3}

Step 3 — Find the terms

We have two possible values for the common ratio.

Case 1: If r=−3/4r = \mathbf{-3/4} and a=−1a = \mathbf{-1}. The terms are a/ra/r, aa, arar.

a/r=−1/(−3/4)=4/3a/r = -1 / (-3/4) = 4/3

a=−1a = -1

ar=−1×(−3/4)=3/4ar = -1 \times (-3/4) = 3/4

The terms are 4/3\mathbf{4/3}, −1\mathbf{-1}, 3/4\mathbf{3/4}.

Case 2: If r=−4/3r = \mathbf{-4/3} and a=−1a = \mathbf{-1}. The terms are a/ra/r, aa, arar.

a/r=−1/(−4/3)=3/4a/r = -1 / (-4/3) = 3/4

a=−1a = -1

ar=−1×(−4/3)=4/3ar = -1 \times (-4/3) = 4/3

The terms are 3/4\mathbf{3/4}, −1\mathbf{-1}, 4/3\mathbf{4/3}.

Answer

(i) The common ratio is −3/4\mathbf{-3/4} or −4/3\mathbf{-4/3}. (ii) The terms are 4/3,−1,3/4\mathbf{4/3, -1, 3/4} or 3/4,−1,4/3\mathbf{3/4, -1, 4/3}.

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