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:

Let $3, 6, 9, 12, \ldots$ up to $78$ terms and $5, 9, 13, 17, \ldots$ up to $59$ terms be two series. Then,the sum of the terms common to both the series is equal to

If $a, b, c$ are in $A.P.$,then $\frac{a}{bc}, \frac{1}{c}, \frac{2}{b}$ are in

The product of three consecutive terms of a $G.P.$ is $512$. If $4$ is added to each of the first and the second of these terms,the three terms now form an $A.P.$ Then the sum of the original three terms of the given $G.P.$ is

If $x = \sum_{n=0}^{\infty} (-1)^{n} \tan^{2n} \theta$ and $y = \sum_{n=0}^{\infty} \cos^{2n} \theta$ for $0 < \theta < \frac{\pi}{4}$,then:

If $|x| < 1, |y| < 1$ and $x \neq y,$ then the sum to infinity of the following series $(x+y)+(x^{2}+xy+y^{2})+(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots$ is: