(a) If (3 - x), 6, (7 - 5x) are consecutive terms of a geometric progression (GP) with constant ratio r > 0, find the : (i) values of x ; (ii) constant rati...

Answer Details

(a) Geometric progression

For three consecutive GP terms, the square of the middle term equals the product of the outer terms:

\[6^2 = (3 - x)(7 - 5x)\]\[36 = 21 - 15x - 7x + 5x^2\]\[36 = 5x^2 - 22x + 21\]\[5x^2 - 22x - 15 = 0\]

(i) Values of x

Using the quadratic formula with \(a = 5,\; b = -22,\; c = -15\):

\[x = \frac{22 \pm \sqrt{(-22)^2 - 4(5)(-15)}}{2(5)} = \frac{22 \pm \sqrt{484 + 300}}{10} = \frac{22 \pm \sqrt{784}}{10} = \frac{22 \pm 28}{10}\]\[x = 5 \quad\text{or}\quad x = -\frac{6}{10} = -\frac{3}{5}\]

(ii) Constant ratio

The ratio is \(r = \dfrac{6}{3 - x}\).

• If \(x = 5\): \(r = \dfrac{6}{3-5} = \dfrac{6}{-2} = -3\) (negative, rejected since \(r > 0\)).
• If \(x = -\dfrac{3}{5}\): \(r = \dfrac{6}{3 + \frac{3}{5}} = \dfrac{6}{\frac{18}{5}} = 6\times\dfrac{5}{18} = \dfrac{5}{3}\) (positive, accepted).

So the admissible value is \(x = -\dfrac{3}{5}\) with constant ratio \(r = \dfrac{5}{3}\). (Check: terms are \(3.6,\; 6,\; 10\), and \(\tfrac{6}{3.6} = \tfrac{10}{6} = \tfrac{5}{3}\).)

(b) Finding \(\angle ADC\)

In the quadrilateral, \(|AB| = 3\text{ cm}\), \(|BC| = 4\text{ cm}\), \(|CD| = 6\text{ cm}\), \(|DA| = 7\text{ cm}\), and the angle at B is \(90^\circ\). Draw the diagonal \(AC\).

In right triangle \(ABC\):

\[|AC|^2 = |AB|^2 + |BC|^2 = 3^2 + 4^2 = 9 + 16 = 25 \;\Rightarrow\; |AC| = 5\text{ cm}\]

In triangle \(ACD\), apply the cosine rule with \(\angle ADC\) opposite \(AC\):

\[|AC|^2 = |DA|^2 + |DC|^2 - 2\,|DA|\,|DC|\cos(\angle ADC)\]\[25 = 7^2 + 6^2 - 2(7)(6)\cos(\angle ADC)\]\[25 = 49 + 36 - 84\cos(\angle ADC)\]\[84\cos(\angle ADC) = 85 - 25 = 60\]\[\cos(\angle ADC) = \frac{60}{84} = 0.7143\]\[\angle ADC = \cos^{-1}(0.7143) = 44.4^\circ \approx 44^\circ\]

Therefore \(\angle ADC \approx 44^\circ\).