(a) Given that $(7 - 2x)$, $9$, $(5x + 17)$ are consecutive terms of a Geometric Progression (G.P.) with a common ratio $r > 0$, we need to find the values of $x$. (b) Two positive numbers are in the ratio $3:4$. The sum of thrice the first number and twice the second is 68. Find the smaller number.

1. Problem Description

(a) Given that $(7 - 2x)$, $9$, $(5x + 17)$ are consecutive terms of a Geometric Progression (G.P.) with a common ratio $r > 0$, we need to find the values of $x$.

(b) Two positive numbers are in the ratio $3:4$. The sum of thrice the first number and twice the second is

6

8. Find the smaller number.

2. Solution Steps

(a)

In a geometric progression, the ratio between consecutive terms is constant. Thus,

$\frac{9}{7 - 2x} = \frac{5x + 17}{9}$

Cross-multiplying, we have

$81 = (7 - 2x)(5x + 17)$

$81 = 35x + 119 - 10x^2 - 34x$

$81 = x + 119 - 10x^2$

$10x^2 - x - 38 = 0$

Using the quadratic formula, we have

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{1 \pm \sqrt{(-1)^2 - 4(10)(-38)}}{2(10)}$

$x = \frac{1 \pm \sqrt{1 + 1520}}{20}$

$x = \frac{1 \pm \sqrt{1521}}{20}$

$x = \frac{1 \pm 39}{20}$

$x_1 = \frac{1 + 39}{20} = \frac{40}{20} = 2$

$x_2 = \frac{1 - 39}{20} = \frac{-38}{20} = -\frac{19}{10} = -1.9$

If $x = 2$, the terms are $7 - 2(2) = 3$, $9$, $5(2) + 17 = 27$. The common ratio is $9/3 = 3$ and $27/9 = 3$, which satisfies $r > 0$.

If $x = -1.9$, the terms are $7 - 2(-1.9) = 7 + 3.8 = 10.8$, $9$, $5(-1.9) + 17 = -9.5 + 17 = 7.5$. The common ratio is $9/10.8 = 0.833...$ and $7.5/9 = 0.833...$. This also satisfies $r > 0$.

(b)

Let the two positive numbers be $3k$ and $4k$, where $k > 0$.

According to the problem, $3(3k) + 2(4k) = 68$.

$9k + 8k = 68$

$17k = 68$

$k = \frac{68}{17} = 4$

The numbers are $3k = 3(4) = 12$ and $4k = 4(4) = 16$.

The smaller number is $12$.

3. Final Answer

(a) $x = 2$ or $x = -\frac{19}{10}$

(b) $12$