## Task B-4.2.

The radius of the base, the slant height, and the height of a right circular cone are in some order three consecutive terms of an increasing arithmetic sequence. If the area of the axial section of the cone is $300 \mathrm{~cm}^{2}$, what is its volume?
## Solution.

If the radius, slant height, and height form an arithmetic sequence with a common difference $d$, there are two possible cases:

1. case: $r=a_{1}, v=a_{1}+d, s=a_{1}+2 d$
2. case: $v=a_{1}, r=a_{1}+d, s=a_{1}+2 d$

First case:

According to the property of an arithmetic sequence, it holds that $d=v-r$ and $s=a_{1}+2 d=r+2(v-r)=2 v-r$.

The area of the base section of the cone is an isosceles triangle and it holds that $P=\frac{1}{2} \cdot 2 r \cdot v$, that is, $r v=300$.

For the slant height, according to the Pythagorean theorem, it holds that $s^{2}=r^{2}+v^{2}$, so $(2 v-r)^{2}=r^{2}+v^{2}$, that is:

$3 v^{2}=4 r v=4 \cdot 300$.

From this, it follows that $v=20 \mathrm{~cm}$ and $r=15 \mathrm{~cm}$, so the desired volume is:

$V_{1}=\frac{1}{3} r^{2} \pi v=1500 \pi \mathrm{cm}^{3}$.

1 point

Second case:

According to the property of an arithmetic sequence, it holds that $d=r-v$ and $s=a_{1}+2 d=v+2(r-v)=2 r-v$.

The area of the base section of the cone is an isosceles triangle and it holds that $P=\frac{1}{2} \cdot 2 r \cdot v$, that is, $r v=300$.

For the slant height, according to the Pythagorean theorem, it holds that $s^{2}=r^{2}+v^{2}$, so $(2 r-v)^{2}=r^{2}+v^{2}$, that is:

$3 r^{2}=4 r v=4 \cdot 300$.

From this, it follows that $r=20 \mathrm{~cm}$ and $v=15 \mathrm{~cm}$, so the desired volume is:

$V_{2}=\frac{1}{3} r^{2} \pi v=2000 \pi \mathrm{cm}^{3}$.