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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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If the $3^{rd}$ term of a GP is 18 and the $10^{th}$ term is 39366 , find the $7^{th}$ term of the GP.

$(a)\;486\qquad(b)\;2187\qquad(c)\;3936\qquad(d)\;1458$

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1 Answer

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Answer : (d) 1458
Explanation :
$T_{3}=ar^2=18$
$T_{10}=ar^9=39366$
$\frac{T_{10}}{T_{3}}=\frac{ar^9}{ar^2}=r^7=\frac{39366}{18}$
$r^7=2187$
$r=3\quad\;,ar^2=18\quad\;,a=18/9=2$
$T_{7}=ar^6=2*3^6=1458.$
answered Jan 3, 2014 by yamini.v
 

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