Gujarati
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

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