Question

If $x_1,x_2,x_3$.......... are ion $G.P$. of natural numbers such that $x_1.x_2.x_3{x}_4=64$ then $x_5=4^k$ find $k$

Answer 1

$\begin{array}{cc}{x}_1,x_2,x_{3}areinG.P& \\ \Rightarrow x_1.x_2.x_3.x_4=64& \\ \Rightarrow x_1.x_2.x_3.x_4.x_5=4^3{.4}^k& \\ let\frac{1}{r^2}.\frac{1}{r}.1,r,r^{2}areinG.P& \\ Hence,\frac{1}{r^2}.\frac{1}{r}.1.r.r^2=4^{3+k}& \\ \Rightarrow 1=4^{3+k}& \left[\begin{array}{c}{a}^x=y\\ x=y\end{array}\right]\\ \Rightarrow 4^0=4^{3+k}& \\ \therefore 3+k=0& \\ k=-3& \end{array}$