Three numbers are in G.P. such that their sum is 42 and their product is 1728. The term(s) of the G.P. is/are?

Steps to Solve

1. Define the terms of the G.P.

Let the three numbers in the geometric progression be $a$, $ar$, and $ar^2$, where $a$ is the first term and $r$ is the common ratio.

2. Set up the sum equation

According to the problem, their sum is 42: $$ a + ar + ar^2 = 42 $$

Factoring out $a$ gives: $$ a(1 + r + r^2) = 42 $$

3. Set up the product equation

The problem states that their product is 1728: $$ a \cdot ar \cdot ar^2 = 1728 $$ This simplifies to: $$ a^3 r^3 = 1728 $$

4. Express variables in terms of each other

From the product equation, we can solve for $a^3$: $$ a^3 = \frac{1728}{r^3} $$

5. Substitute into the sum equation

Substituting $a^3$ into the sum equation gives: $$ \sqrt[3]{\frac{1728}{r^3}}(1 + r + r^2) = 42 $$

6. Solve for $r$

Calculating $1728$ gives: $$ 1728 = 12^3 \implies a = 12/r $$ So substituting: $$ (1 + r + r^2) \cdot \frac{12}{r} = 42 $$ This simplifies to: $$ 12(1 + r + r^2) = 42r $$

7. Rearranging the equation

Rearranging it leads to: $$ 12 + 12r + 12r^2 = 42r $$ $$ 12r^2 - 30r + 12 = 0 $$

8. Using the quadratic formula

Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 12$, $b = -30$, and $c = 12$: $$ r = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 12 \cdot 12}}{2 \cdot 12} $$ $$ r = \frac{30 \pm \sqrt{900 - 576}}{24} $$ $$ r = \frac{30 \pm \sqrt{324}}{24} $$ $$ r = \frac{30 \pm 18}{24} $$

Calculating gives two possible values: $$ r_1 = 2 \quad \text{and} \quad r_2 = \frac{1}{3} $$

9. Find corresponding values of a

For $r = 2$: $$ a = \frac{12}{2} = 6 $$ Numbers are $6$, $12$, $24$.

For $r = \frac{1}{3}$: $$ a = 12 \cdot 3 = 36 $$ Numbers are $36$, $12$, $4$.

10. Verify the sums and products

Verify both sets:

1. For $6, 12, 24$:

   • Sum: $6 + 12 + 24 = 42$, Product: $6 \cdot 12 \cdot 24 = 1728$.

2. For $36, 12, 4$:

   • Sum: $36 + 12 + 4 = 52$, Product: $36 \cdot 12 \cdot 4 = 1728$ (not valid).

Final valid numbers in G.P. are: $$ 6, 12, 24 $$.