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8. Sequences and Series
easy
The sum of $3$ numbers in geometric progression is $38$ and their product is $1728$. The middle number is
A
$12$
B
$8$
C
$18$
D
$6$
Solution
(a) Let the three numbers be $\frac{a}{r},\;a$ and $ar$, then
${a^3} = 1728$
$ \Rightarrow $ $a = 12$
Hence the middle term is $12.$
Standard 11
Mathematics