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Let $d(n)$ denotes the number of divisors of $n$ including 1 and itself. Then, $d(225), d(1125)$ and $d(640)$ are
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The correct answer is:
in GP
Here, $225=3^{2} \times 5^{2}$
$\Rightarrow d(225)=3 \times 3=9$
$1126=3^{2} \times 5^{3}$
$\Rightarrow \quad d(1125)=3 \times 4=12$
and $640=2^{7} \times 5$
$\Rightarrow \quad d(640)=8 \times 2=16$
Hence, 9,12 and 16 are in GP.
$\Rightarrow d(225)=3 \times 3=9$
$1126=3^{2} \times 5^{3}$
$\Rightarrow \quad d(1125)=3 \times 4=12$
and $640=2^{7} \times 5$
$\Rightarrow \quad d(640)=8 \times 2=16$
Hence, 9,12 and 16 are in GP.
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