NIMCET 2019 Question Paper with solutions

Concept:
The sides of an obtuse triangle should satisfy the condition that the sum of the square of two smaller sides should be less than the square of the largest side.
If $\mathit{a},\mathit{b},\mathit{c}$ are in $\mathit{A}\mathit{P}$ than $\mathit{b}=\frac{\mathit{a}+\mathit{c}}{2}$
If $\mathit{a},\mathit{b},\mathit{c}$ are in GP than $\mathit{b}=\sqrt{\mathit{a}\mathit{c}}$
Calculation:
$\log \mathit{a}-\log 2\mathit{b},\log 2\mathit{b}-\log 3\mathit{c}$ and $\log 3\mathit{c}-\log$ a are in AP
$\log 2\mathit{b}-\log 3\mathit{c}=\frac{\log 3\mathit{c}-\log \mathit{a}+\log \mathit{a}-\log 2\mathit{b}}{2}$
$\log \frac{2\mathit{b}}{3\mathit{c}}=\frac{\log \frac{3\mathit{c}}{2\mathit{b}}}{2}$
$\frac{4{\mathit{b}}^2}{9{\mathit{c}}^2}=\frac{3\mathit{c}}{2\mathit{b}}$
$2\mathit{b}=3\mathit{c}$-⋅s....(1)
$\because \mathit{a},\mathit{b},\mathit{c}$ are in GP
$\Rightarrow {\mathit{b}}^2=\mathit{a}\mathit{c}...\left(2\right)$
From equation 1 and 2 we get
$9\mathit{c}=4\mathit{a}$
$\mathit{a}:\mathit{b}:\mathit{c}=9:2:4$
As we know that, for obtuse triangle the sum of the square of two smaller sides should be less than the square of the largest side.
$\Rightarrow {\mathit{a}}^2>{\mathit{b}}^2+{\mathit{c}}^2$
So, $\mathit{a},\mathit{b}$ and $\mathit{c}$ are the sides of an obtuse angle triangle.