Question

Sol. [Key Concept: G.P. with first four terms summing to 15 ⇒ solve for $a,r$.]

$$a\left( 1 + r + r^{2} + r^{3} \right) = 15$$

$$\begin{matrix} & a^{2}\left( 1 + r^{2} + r^{4} + r^{6} \right) = 85 \Rightarrow \ \frac{\left( 1 + r + r^{2} + r^{3} \right)^{2}}{\left( 1 + r^{2} + r^{4} + r^{6} \right)} = \frac{225}{85} \\ & \ \Rightarrow \ \frac{\left( 1 - r^{4} \right)^{2}\left( 1 - r^{2} \right)}{(1 - r)^{2}\left( 1 - r^{8} \right)} = \frac{45}{17} = \frac{\left( 1 - r^{4} \right)(1 + r)}{\left( 1 + r^{4} \right)(1 - r)} \end{matrix}$$

$$\begin{matrix} \Rightarrow & 14r^{4} - 17r^{3} - 17r^{2} - 17r + 14 = 0 \\ \Rightarrow & 14\left( r^{2} + \frac{1}{r^{2}} \right) - 17\left( r + \frac{1}{r} \right) - 17 = 0 \\ \Rightarrow & 14\left( t^{2} - 2 \right) - 17t - 17 = 0\ t = r + \frac{1}{r} \\ \Rightarrow & 14t^{2} - 17t - 45 = 0 = (7t + 9)(2t - 5) \\ \Rightarrow & r + \frac{1}{r} = \frac{5}{2}\ \Rightarrow \ r = \frac{1}{2},2 \\ & r = \frac{1}{2},a = 8\ \text{~}\text{or}\text{~}\ r = 2,a = 1 \end{matrix}$$

✔ (a) $r = 2,a = 1 \Rightarrow \ t_{6 = 32}$

✘ (b) not a sixth term

✔ (c) $r = \frac{12,}{a} = 8 \Rightarrow \ t_{6 =}\frac{14}{}$

✘ (d) not a sixth term