Question

if the nth term of a geometric progression 162, 54, 18,.... and 2/81, 2/27, 2/9 ... are qqual, then find the value of n
pls answer me as soon as possible..

Answer 1

n=5
for a geometric progression, the nth term is a(r)^n-1. a is the first term. r is the common ratio.
so,
A(R)^n-1 = a(r)^n-1
A=162, R=54/162 for the 1st geometric progression.
a=2/81, r= (2/27)/(2/81) for the 2nd geometric progression.
solve and get the value for n.

hope this helped!
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Answer 2

There are two geometric progressions. 
The nth term of both are same.

for 162,54,18...
a = 162, r = 54/162 = 1/3
for 2/81, 2/27, 2/9...
a = 2/81, r = (2/27) / (2/81) = 81/27 = 3

$162 \times \frac{1}{3^{n-1}} = \frac{2}{81} \times 3^{n-1}\\ \\ \frac{162}{3^{n-1}} = \frac{2 \times 3^{n-1}}{81}\\ \\2 \times 3^{2(n-1)}=162 \times 81 \\ \\2 \times 3^{2(n-1)}=2 \times 81 \times 81 \\ \\2 \times 3^{2(n-1)}=2 \times 3^4 \times 3^4=2 \times 3^8\\ \\3^{2(n-1)}= 3^8\\ \\2(n-1)=8\\ \\n-1= \frac{8}{2}=4\\ \\n=4+1=\boxed{5}$