Question

In the geometric progression 16807,2401,343.....and 1/243,1/81,1/27,... nth terms are same then find n​

Answer 1

Given:

$16807,2401,343 .... \ \ \ \ and \ \ \ \ \frac{1}{243},\frac{1}{81},\frac{1}{27}..... \ in \ G.P$

To find:

n= ?

when nth terms are same

Solution:

In series 16807,2401,343:

$a= 16807 \\\\r= \frac{2401}{16807} \\\\r= 0.14$

In series

$\frac{1}{243},\frac{1}{81},\frac{1}{27}:\\\\a= \frac{1}{243}\\\\r= \frac{\frac{1}{81}}{\frac{1}{243}}\\\\r= \frac{1}{81}} \times \frac{243}{1}}\\\\r= 3$

Formula:

$T_n= ar^{n-1}\\\\when \ n= n\\\\ar^{n-1}=ar^{n-1}\\\\\Rightarrow 16807 \times (0.14)^{n-1}= \frac{1}{243} \times (3)^{n-1}\\\\ \Rightarrow 16087 \times 243 \times (0.14)^{n}\times (0.14)^{-1}=3^n\times 3^{-1}\\\\\Rightarrow 16087 \times 243 \times (0.14)^{n}\times \frac{1}{0.14}=3^n\times \frac{1}{3}\\\\\Rightarrow \frac{16087 \times 243 \times 3 }{0.14}= \frac{3^n}{0.14^n}\\\\\Rightarrow 83,767,307.1 = \frac{3^n}{0.14^n}$