Question 13

Let's find the three terms of the geometric progression.

Step 1 — Set up equations

Let's call the three terms $a$, $ar$, and $ar^2$. We write down the first condition. The sum of the terms is 26.

$a + ar + ar^2 = 26$

We can factor out $a$ from the left side.

$a(1 + r + r^2) = 26 \quad \text{(Equation 1)}$

Now, we write down the second condition. The sum of their squares is 364.

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

We square each term.

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

We can factor out $a^2$ from the left side.

$a^2(1 + r^2 + r^4) = 364 \quad \text{(Equation 2)}$

Step 2 — Find the common ratio

Now, we divide Equation 2 by the square of Equation 1.

$\frac{a^2(1 + r^2 + r^4)}{[a(1 + r + r^2)]^2} = \frac{364}{26^2}$

We simplify the left side. We also calculate $26^2$.

$\frac{a^2(1 + r^2 + r^4)}{a^2(1 + r + r^2)^2} = \frac{364}{676}$

We cancel $a^2$ from the top and bottom. We simplify the fraction on the right.

$\frac{1 + r^2 + r^4}{(1 + r + r^2)^2} = \frac{7}{13}$

We use a special algebraic identity: $1 + r^2 + r^4 = (1 + r + r^2)(1 - r + r^2)$. We substitute this into our equation.

$\frac{(1 + r + r^2)(1 - r + r^2)}{(1 + r + r^2)^2} = \frac{7}{13}$

We cancel one $(1 + r + r^2)$ term from the top and bottom.

$\frac{1 - r + r^2}{1 + r + r^2} = \frac{7}{13}$

We cross-multiply to remove fractions.

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

We distribute the numbers on both sides.

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

We rearrange the terms to form a quadratic equation.

$13r^2 - 7r^2 - 13r - 7r + 13 - 7 = 0$

We combine like terms.

$6r^2 - 20r + 6 = 0$

We divide the entire equation by 2 to simplify it.

$3r^2 - 10r + 3 = 0$

We solve this quadratic equation for $r$ by factoring.

$3r^2 - 9r - r + 3 = 0$

$3r(r - 3) - 1(r - 3) = 0$

$(3r - 1)(r - 3) = 0$

This gives us two possible values for $r$.

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

$\boxed{r = \frac{1}{3} \quad \text{or} \quad r = 3}$

Step 3 — Find the terms

We consider the first possible value for $r$. If $r = \frac{1}{3}$. We substitute this value into Equation 1: $a(1 + r + r^2) = 26$.

$a\left(1 + \frac{1}{3} + \left(\frac{1}{3}\right)^2\right) = 26$

$a\left(1 + \frac{1}{3} + \frac{1}{9}\right) = 26$

We find a common denominator for the terms inside the parenthesis.

$a\left(\frac{9}{9} + \frac{3}{9} + \frac{1}{9}\right) = 26$

$a\left(\frac{9 + 3 + 1}{9}\right) = 26$

$a\left(\frac{13}{9}\right) = 26$

We solve for $a$.

$a = 26 \times \frac{9}{13}$

$a = 2 \times 9$

$\boxed{a = 18}$

The three terms are $a$, $ar$, $ar^2$. The terms are 18, $18 \times \frac{1}{3} = \textbf{6}$, and $18 \times \left(\frac{1}{3}\right)^2 = 18 \times \frac{1}{9} = \textbf{2}$. So, the terms are 18, 6, 2.

Now, we consider the second possible value for $r$. If $r = 3$. We substitute this value into Equation 1: $a(1 + r + r^2) = 26$.

$a(1 + 3 + 3^2) = 26$

$a(1 + 3 + 9) = 26$

$a(13) = 26$

We solve for $a$.

$a = \frac{26}{13}$

$\boxed{a = 2}$

The three terms are $a$, $ar$, $ar^2$. The terms are 2, $2 \times 3 = \textbf{6}$, and $2 \times 3^2 = 2 \times 9 = \textbf{18}$. So, the terms are 2, 6, 18.

The terms of the GP are the same in both cases, just in a different order.

Answer

The terms of the GP are 2, 6, 18.