logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
0 votes

Find n such that is $\large\frac{512}{729}\;$is the nth term of the GP 18,-12,8,-----------------

$\begin{array}{1 1} 8 \\ 18 \\ 12 \\ 9 \end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Answer : (d) 9
Explanation :$ a=18 , r=\large\frac{-12}{18}=\large\frac{-2}{3}$
$T_{n}=ar^{n-1}\qquad\;\frac{512}{729}=18*(-2/3)^{(n-1)}$
$(-2/3)^{(n-1)}=\large\frac{512}{18*729}=\large\frac{2^9}{2*3^2*9^3}=\large\frac{2^8}{3^2*3^6}$
$=\large\frac{2^8}{3^8}$
$n-1=8\quad\;n=9.$
answered Jan 1, 2014 by yamini.v
edited Jan 24, 2014
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...