If $a$ is the arithmetic mean of $b$ and $c$ and $G_1, G_2$ are the two geometric means between them,then $G_1^3 + G_2^3 = $

Let $x_1, x_2, x_3, x_4$ be in a geometric progression. If $2, 7, 9, 5$ are subtracted respectively from $x_1, x_2, x_3, x_4$,then the resulting numbers are in an arithmetic progression. Then the value of $\frac{1}{24}(x_1 x_2 x_3 x_4)$ is:

Three unequal positive numbers $a, b, c$ are such that $a, b, c$ are in $G.P.$ while $\log \left(\frac{5 c}{2 a}\right), \log \left(\frac{7 b}{5 c}\right), \log \left(\frac{2 a}{7 b}\right)$ are in $A.P.$ Then $a, b, c$ are the lengths of the sides of

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 $n=1, 2, 3, \ldots$. If $\left(\frac{a_{1}+\frac{1}{a_{2}}}{a_{3}}\right) \cdot\left(\frac{a_{2}+\frac{1}{a_{3}}}{a_{4}}\right) \cdot\left(\frac{a_{3}+\frac{1}{a_{4}}}{a_{5}}\right) \cdots\left(\frac{a_{30}+\frac{1}{a_{31}}}{a_{32}}\right)=2^{\alpha}\left({}^{61}C_{31}\right)$,then $\alpha$ is equal to.

If $9, x, y, z, a$ are in $A.P.$ such that $x + y + z = 15$,and $9, x, y, z, a$ are in $H.P.$ such that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3}$,then the value of $a$ is:

Write the first five terms of the following sequence and obtain the corresponding series:
$a_{1} = a_{2} = 2, a_{n} = a_{n-1} - 1, n > 2$