In $\triangle \mathrm{ABC}$, if a, b, c are in arithmetic progression and $\mathrm{C}=2 \mathrm{~A}$, then…

In $\triangle \mathrm{ABC}$, if a, b, c are in arithmetic progression and $\mathrm{C}=2 \mathrm{~A}$, then $\mathrm{a}: \mathrm{c}=$
  1. $4: 5$
  2. $2: 3$
  3. $5: 6$
  4. $\sqrt{3}: 2$

Solution

Given that $a, b, c$ are in A.P $\begin{aligned} & \Rightarrow \mathrm{b}=\frac{\mathrm{a}+\mathrm{c}}{2} \ldots \text { (i) } \\ & \text { And } \mathrm{C}=2 \mathrm{~A} \Rightarrow \sin \mathrm{C}=\sin 2 \mathrm{~A}=2 \sin \mathrm{A} \cdot \cos \mathrm{A} \\ & \Rightarrow \mathrm{C}=2 \mathrm{a} \cdot \frac{\left(\mathrm{b}^2+\mathrm{c}^2-\mathrm{a}^2\right)}{2 \mathrm{bc}} \\ & \Rightarrow 2 \mathrm{c}^3+3 \mathrm{a}^3-3 \mathrm{ac}^2-2 \mathrm{a}^2 \mathrm{c}=0 \\ & \Rightarrow 3\left(\frac{\mathrm{a}}{\mathrm{c}}\right)^3-2\left(\frac{\mathrm{a}}{\mathrm{c}}\right)^2-3\left(\frac{\mathrm{a}}{\mathrm{c}}\right)+2=0 \\ & \Rightarrow\left(3\left(\frac{\mathrm{a}}{\mathrm{c}}\right)-2\right)\left[\left(\frac{\mathrm{a}}{\mathrm{c}}\right)^2-1\right]=0 \Rightarrow \frac{\mathrm{a}}{\mathrm{c}}=\frac{2}{3}=2: 3 \end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 2)

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