The first term of an arithmetic progression is $1$. If the second,tenth,and thirty-fourth terms form a geometric progression,then the common difference of the arithmetic progression is:

Let $0 < z < y < x$ be three real numbers such that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an arithmetic progression and $x, \sqrt{2}y, z$ are in a geometric progression. If $xy + yz + zx = \frac{3}{\sqrt{2}} xyz$,then $3(x + y + z)^2$ is equal to $............$.

The sum of two numbers is $6$ times their geometric mean. Show that the numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$.

Three non-zero real numbers form an $A.P.$ and the squares of these numbers taken in the same order form a $G.P.$ Then the number of all possible common ratios of the $G.P.$ is

Let $b_i > 1$ for $i = 1, 2, \ldots, 101$. Suppose $\log _e b_1, \log _e b_2, \ldots, \log _e b_{101}$ are in Arithmetic Progression $(A.P.)$ with the common difference $\log _e 2$. Suppose $a_1, a_2, \ldots, a_{101}$ are in $A.P.$ such that $a_1 = b_1$ and $a_{51} = b_{51}$. If $t = b_1 + b_2 + \cdots + b_{51}$ and $s = a_1 + a_2 + \cdots + a_{51}$,then:

The common difference of an $A.P.$ whose first term is unity and whose second,tenth and thirty-fourth terms are in $G.P.$ is