Insert three numbers between $1$ and $256$ so that the resulting sequence is a $G.P.$
Let $G_{1}, G_{2}, G_{3}$ be three numbers between $1$ and $256$ such that $1, G _{1}, G _{2}, G _{3}, 256$ is a $G.P.$
Therefore $\quad 256=r^{4}$ giving $r=\pm 4$ (Taking real roots only)
For $r=4,$ we have $G _{1}=a r=4, G _{2}=a r^{2}=16, G _{3}=a r^{3}=64$
Similarly, for $r=-4,$ numbers are $-4,16$ and $-64$ Hence, we can insert $4,16,64$ between $1$ and $256$ so that the resulting sequences are in $G.P.$
The first term of an infinite geometric progression is $x$ and its sum is $5$. Then
Find the sum of first $n$ terms and the sum of first $5$ terms of the geometric
series $1+\frac{2}{3}+\frac{4}{9}+\ldots$
If $x > 1,\;y > 1,z > 1$ are in $G.P.$, then $\frac{1}{{1 + {\rm{In}}\,x}},\;\frac{1}{{1 + {\rm{In}}\,y}},$ $\;\frac{1}{{1 + {\rm{In}}\,z}}$ are in
Sum of infinite number of terms in $G.P.$ is $20$ and sum of their square is $100$. The common ratio of $G.P.$ is
Given $a_1,a_2,a_3.....$ form an increasing geometric progression with common ratio $r$ such that $log_8a_1 + log_8a_2 +.....+ log_8a_{12} = 2014,$ then the number of ordered pairs of integers $(a_1, r)$ is equal to