Predicting What Comes Next: Sequences and Progressions | EOT

Question 13

The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.

Solution

Understand the Question

Let the first three terms of the Geometric Progression (GP) be $a$ , $ar$ , and $ar^2$ , where $a$ is the first term and $r$ is the common ratio.
We are given two conditions:
1. Sum of the terms: $a + ar + ar^2 = 26$
2. Sum of their squares: $a^2 + (ar)^2 + (ar^2)^2 = 364$
We use the algebraic identity $1 + r^2 + r^4 = (1 + r + r^2)(1 - r + r^2)$ to solve for $r$ , and then find the corresponding values of $a$ to determine the terms.

Step 1 · Set up the Equations

Let the first three terms of the GP be $a$ , $ar$ , and $ar^2$ . Diagram 1

From the first condition

a + ar + ar^2 = 26

a(1 + r + r^2) = 26 \quad \dots (1)

From the second condition

(a)^2 + (ar)^2 + (ar^2)^2 = 364

a^2 + a^2r^2 + a^2r^4 = 364

a^2(1 + r^2 + r^4) = 364 \quad \dots (2)

Step 2 · Find the Common Ratio $r$

Divide Equation $(2)$ by the square of Equation $(1)$

\begin{aligned} \dfrac{a^2(1 + r^2 + r^4)}{[a(1 + r + r^2)]^2} &= \dfrac{364}{26^2} \\[1.1em] \dfrac{a^2(1 + r^2 + r^4)}{a^2(1 + r + r^2)^2} &= \dfrac{364}{676} \\[1.1em] \dfrac{1 + r^2 + r^4}{(1 + r + r^2)^2} &= \dfrac{7}{13} \end{aligned}

Using the identity $1 + r^2 + r^4 = (1 + r + r^2)(1 - r + r^2)$

\begin{aligned} \dfrac{(1 + r + r^2)(1 - r + r^2)}{(1 + r + r^2)^2} &= \dfrac{7}{13} \\[1.1em] \dfrac{1 - r + r^2}{1 + r + r^2} &= \dfrac{7}{13} \end{aligned}

Cross-multiplying

13(1 - r + r^2) = 7(1 + r + r^2)

13 - 13r + 13r^2 = 7 + 7r + 7r^2

\begin{aligned} 13r^2 - 7r^2 - 13r - 7r + 13 - 7 &= 0 \\[0.6em] 6r^2 - 20r + 6 &= 0 \\[0.6em] 3r^2 - 10r + 3 &= 0 \end{aligned}

Factoring the quadratic equation

\begin{aligned} 3r^2 - 9r - r + 3 &= 0 \\[0.6em] 3r(r - 3) - 1(r - 3) &= 0 \\[0.6em] (3r - 1)(r - 3) &= 0 \end{aligned}

3r - 1 = 0 \quad \text{or} \quad r - 3 = 0

r = \dfrac{1}{3} \quad \text{or} \quad r = 3

Step 3 · Find the Terms of the GP

Case 1: When $r = \dfrac{1}{3}$

Substitute into Equation $(1)$

\begin{aligned} a\left(1 + \dfrac{1}{3} + \left(\dfrac{1}{3}\right)^2\right) &= 26 \\[1.1em] a\left(1 + \dfrac{1}{3} + \dfrac{1}{9}\right) &= 26 \\[1.1em] a\left(\dfrac{9 + 3 + 1}{9}\right) &= 26 \\[1.1em] a\left(\dfrac{13}{9}\right) &= 26 \\[1.1em] a &= 26 \times \dfrac{9}{13} \\[0.6em] a &= 18 \end{aligned}

The three terms are $a, ar, ar^2$ : $18, \quad 18 \times \dfrac{1}{3} = 6, \quad 18 \times \dfrac{1}{9} = 2$

Case 2: When $r = 3$

Substitute into Equation $(1)$

\begin{aligned} a(1 + 3 + 3^2) &= 26 \\[0.6em] a(1 + 3 + 9) &= 26 \\[0.6em] a(13) &= 26 \\[0.6em] a &= \dfrac{26}{13} = 2 \end{aligned}

The three terms are $a, ar, ar^2$ : $2, \quad 2 \times 3 = 6, \quad 2 \times 3^2 = 18$

Both cases give the same set of numbers: $2, 6, 18$ .

Answer

$2, 6, 18$ (or $18, 6, 2$ )

Common Mistakes

Identity Error: Forgetting the identity $1 + r^2 + r^4 = (1 + r + r^2)(1 - r + r^2)$ when simplifying $\dfrac{1 + r^2 + r^4}{(1 + r + r^2)^2}$ .
Missing a Solution: Discarding $r = \dfrac{1}{3}$ or $r = 3$ . Both values are valid and yield the same set of terms in ascending and descending order.
Squaring Error: Squaring $ar^2$ as $ar^4$ instead of $(ar^2)^2 = a^2r^4$ .

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