In a G.P.,the 3^{rd} term is 24 and the 6^{th | vedclass

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(A) Let the first term be $a$ and the common ratio be $r$.In a $G.P.$,the $n^{th}$ term is given by $a_n = ar^{n-1}$.Given $a_3 = ar^2 = 24$  $(1)$.Gi

Vedclass · Accepted Answer

$3072$

Vedclass · Answer

(A) Let the first term be $a$ and the common ratio be $r$.In a $G.P.$,the $n^{th}$ term is given by $a_n = ar^{n-1}$.Given $a_3 = ar^2 = 24$  $(1)$.Given $a_6 = ar^5 = 192$  $(2)$.Dividing $(2)$ by $(1)$,we get $\frac{ar^5}{ar^2} = \frac{192}{24}$.$r^3 = 8$,which implies $r = 2$.Substituting $r = 2$ in $(1)$,we get $a(2)^2 = 24$,so $4a = 24$,which gives $a = 6$.The $10^{th}$ term is $a_{10} = ar^9 = 6 	imes (2)^9$.$a_{10} = 6 	imes 512 = 3072$.

In a $G.P.$,the $3^{rd}$ term is $24$ and the $6^{th}$ term is $192$. Find the $10^{th}$ term.

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