If $\alpha, \beta, \gamma$ are the roots of the equation $3 x^3-26 x^2+52 x-24=0$ such that $\alpha, \beta, \gamma$ are in geometric progression and $\alpha<\beta<\gamma$, then $3 \alpha+2 \beta+\gamma=$

Let $f(n)=A(-2)^n+B(-3)^n \forall A, B \in \mathbf{R}$ and $n \in \mathbf{N}-\{1,2\}$. If $f(n)+a f(n-1)+b f(n-2)=0$, then $(a+b)(b-a)=$

If $1+\frac{\cos \theta}{2}+\frac{\cos 2 \theta}{4}+\frac{\cos 3 \theta}{8}+\ldots \ldots=\frac{a-2 \cos \theta}{5+b \cos \theta}$ for some $a, b \in \mathbf{R}$, then $(a-b)^2=$

If $f(1)=3$, and $f(n+1)-f(n)=3\left(4^n-1\right)$, then $\forall n \in \mathbf{N}$, $f(n)=$