$$ \begin{aligned} &\text { Let }\left\{a_{n}\right\}_{n=0}^{\infty} \text { be a sequence such that } a_{0}=a_{1}=0 \text { and } \\\\ &a_{n+2}=3 a_{n+1}-2 a_{n}+1, \forall n \geq 0 . \end{aligned} $$

Then $$a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24}$$ is equal to

Consider the sequence $$a_{1}, a_{2}, a_{3}, \ldots$$ such that $$a_{1}=1, a_{2}=2$$ and $$a_{n+2}=\frac{2}{a_{n+1}}+a_{n}$$ for $$\mathrm{n}=1,2,3, \ldots .$$ If $$\left(\frac{\mathrm{a}_{1}+\frac{1}{\mathrm{a}_{2}}}{\mathrm{a}_{3}}\right) \cdot\left(\frac{\mathrm{a}_{2}+\frac{1}{\mathrm{a}_{3}}}{\mathrm{a}_{4}}\right) \cdot\left(\frac{\mathrm{a}_{3}+\frac{1}{\mathrm{a}_{4}}}{\mathrm{a}_{5}}\right) \ldots\left(\frac{\mathrm{a}_{30}+\frac{1}{\mathrm{a}_{31}}}{\mathrm{a}_{32}}\right)=2^{\alpha}\left({ }^{61} \mathrm{C}_{31}\right)$$, then $$\alpha$$ is equal to :

Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5 . Let the sum of its first five terms be $$\frac{98}{25}$$. Then the sum of the first 21 terms of an AP, whose first term is $$10\mathrm{a r}, \mathrm{n}^{\text {th }}$$ term is $$\mathrm{a}_{\mathrm{n}}$$ and the common difference is $$10 \mathrm{ar}^{2}$$, is equal to :

Suppose $$a_{1}, a_{2}, \ldots, a_{n}$$, .. be an arithmetic progression of natural numbers. If the ratio of the sum of first five terms to the sum of first nine terms of the progression is $$5: 17$$ and , $$110 < {a_{15}} < 120$$, then the sum of the first ten terms of the progression is equal to