We are given two math problems. (a) The terms $(7-2x)$, $9$, and $(5x+17)$ are consecutive terms of a geometric progression (G.P.) with a common ratio $r>0$. We need to find the value(s) 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. We need to find the smaller number.

1. Problem Description

We are given two math problems.

(a) The terms $(7-2x)$, $9$, and $(5x+17)$ are consecutive terms of a geometric progression (G.P.) with a common ratio $r>0$. We need to find the value(s) 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. We need to find the smaller number.

2. Solution Steps

(a) For a geometric progression, the ratio between consecutive terms is constant. Therefore, we have

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

Cross-multiplying gives

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

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

$$10x^2 + 34x - 119 - 35 + 81 = 0$$

$$10x^2 + 34x - 73 = 0$$

We can solve this quadratic equation using the quadratic formula:

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

$$x = \frac{-34 \pm \sqrt{34^2 - 4(10)(-73)}}{2(10)}$$

$$x = \frac{-34 \pm \sqrt{1156 + 2920}}{20}$$

$$x = \frac{-34 \pm \sqrt{4076}}{20}$$

$$x = \frac{-34 \pm 2\sqrt{1019}}{20}$$

$$x = \frac{-17 \pm \sqrt{1019}}{10}$$

Since $r>0$, we must have $7-2x > 0$ and $5x+17 > 0$.

If $x = \frac{-17 + \sqrt{1019}}{10}$, then $x \approx \frac{-17 + 31.92}{10} \approx 1.492$.

If $x = \frac{-17 - \sqrt{1019}}{10}$, then $x \approx \frac{-17 - 31.92}{10} \approx -4.892$.

If $x \approx 1.492$, $7-2x \approx 7-2.984 = 4.016 > 0$ and $5x+17 \approx 5(1.492) + 17 \approx 7.46+17 = 24.46 > 0$.

If $x \approx -4.892$, $7-2x \approx 7 - 2(-4.892) = 7 + 9.784 = 16.784 > 0$ and $5x+17 \approx 5(-4.892) + 17 = -24.46+17 = -7.46 < 0$. This solution is not valid since $r$ needs to be positive and this results in $r = \frac{5x+17}{9} < 0$.

Thus, $x = \frac{-17 + \sqrt{1019}}{10}$.

(b) Let the two positive numbers be $3k$ and $4k$ for some positive number $k$.

We are given that $3(3k) + 2(4k) = 68$.

So, $9k + 8k = 68$.

$$17k = 68$$

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

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

The smaller number is

1

2.

3. Final Answer

(a) $x = \frac{-17 + \sqrt{1019}}{10}$

(b) 12