Theory of Equations 1 Question 57
58. If $a _4=28$, then $p+2 q=$
(a) 14
(b) 7
(c) 21
(d) 12
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Solution:
- $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2}$
$a _4=a _3+a _2$
$=2 a _2+a _1$
$=3 a _1+2 a _0$
$28=p(3 \alpha+2)+q(3 \beta+2)$
$$ \begin{aligned} & 28=(p+q) \frac{3}{2}+2+(p-q) \frac{3 \sqrt{5}}{2} \\ & \therefore \quad p-q=0 \\ & \text { and } \quad(p+q) \times \frac{7}{2}=28 \\ & \Rightarrow \quad p+q=8 \\ & \Rightarrow \quad p=q=4 \\ & \therefore \quad p+2 q=12 \end{aligned} $$
