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8. Sequences and Series
easy
If five $G.M.’s$ are inserted between $486$ and $2/3$ then fourth $G.M.$ will be
A
$4$
B
$6$
C
$12$
D
$-6$
Solution
(b) Let ${G_1},{G_2},{G_3},{G_4},{G_5}$ be the $G.M.’s$ are inserted between $486$ and $2/3$.
So total terms are $7$.
${T_n} = a{r^{n – 1}}$
==> $2/3 = 486$${(r)^6}$
$ \Rightarrow r = 1/3$
Hence $4^{th}$ $G.M.$ will be,
${T_5} = a{r^4} = 486\,{(1/3)^4} = 6$.
Standard 11
Mathematics