Consider an infinite $G.P. $ with first term a and common ratio $r$, its sum is $4$ and the second term is $3/4$, then
$a = \frac{7}{4},\,r = \frac{3}{7}$
$a = \frac{3}{2},\,r = \frac{1}{2}$
$a = 2,\,r = \frac{3}{8}$
$a = 3,\,r = \frac{1}{4}$
If the sum of the series $1 + \frac{2}{x} + \frac{4}{{{x^2}}} + \frac{8}{{{x^3}}} + ....\infty $ is a finite number, then
If $a _{1}(>0), a _{2}, a _{3}, a _{4}, a _{5}$ are in a G.P., $a _{2}+ a _{4}=2 a _{3}+1$ and $3 a _{2}+ a _{3}=2 a _{4}$, then $a _{2}+ a _{4}+2 a _{5}$ is equal to
If the sum of an infinite $GP$ $a, ar, ar^{2}, a r^{3}, \ldots$ is $15$ and the sum of the squares of its each term is $150 ,$ then the sum of $\mathrm{ar}^{2}, \mathrm{ar}^{4}, \mathrm{ar}^{6}, \ldots$ is :
If the first term of a $G.P.$ be $5$ and common ratio be $ - 5$, then which term is $3125$
Let $a$ and $b$ be roots of ${x^2} - 3x + p = 0$ and let $c$ and $d$ be the roots of ${x^2} - 12x + q = 0$, where $a,\;b,\;c,\;d$ form an increasing G.P. Then the ratio of $(q + p):(q - p)$ is equal to