The solution of the equation $1 + a + a^2 + a^3 + \dots + a^x = (1 + a)(1 + a^2)(1 + a^4)$ is given by $x$ is equal to

The sum of a $G.P.$ with common ratio $r = 3$ is $364$,and the last term is $243$. Find the number of terms $n$.

The geometric series $a + ar + ar^2 + ar^3 + \dots \infty$ has a sum of $7$,and the sum of the terms involving odd powers of $r$ is $3$. Then the value of $(a^2 - r^2)$ is -

If $r > 1$,$x = a + \frac{a}{r} + \frac{a}{r^2} + \dots \infty$,$y = b - \frac{b}{r} + \frac{b}{r^2} - \dots \infty$,and $z = c + \frac{c}{r^2} + \frac{c}{r^4} + \dots \infty$,then $\frac{xy}{z} = \dots$

If $a, b, c, d$ and $p$ are different real numbers such that $(a^{2}+b^{2}+c^{2}) p^{2}-2(ab+bc+cd) p+(b^{2}+c^{2}+d^{2}) \leq 0$,then show that $a, b, c$ and $d$ are in $G.P.$

Let $a_1, a_2, a_3, \ldots$ be a $GP$ of increasing positive numbers. If the product of the fourth and sixth terms is $9$ and the sum of the fifth and seventh terms is $24$,then $a_1 a_9 + a_2 a_4 a_9 + a_5 + a_7$ is equal to $.........$.