**Answer:**

**Given:**
- The frequency of a note emitted by a source changes by 20% as it approaches a stationary observer.
- Velocity of sound in air is 340 m/s.

**To Find:**
- The apparent frequency when it recedes away from the observer.

We know that the change in frequency of sound is given by the Doppler's effect formula:

$f' = \frac{v \pm v_o}{v \pm v_s}f$

where
- $f$ is the actual frequency of the sound wave emitted by the source.
- $f'$ is the apparent frequency of the sound wave received by the observer.
- $v$ is the velocity of sound in air.
- $v_o$ is the velocity of observer.
- $v_s$ is the velocity of source.

**Approach:**
- First, we need to find the velocity of the source and observer when the sound wave approaches the observer.
- Then, we can use the Doppler's effect formula to find the apparent frequency of the sound wave received by the observer.
- Next, we need to find the velocity of the source and observer when the sound wave recedes away from the observer.
- Finally, we can use the same Doppler's effect formula to find the apparent frequency of the sound wave received by the observer.

**Calculation:**

**When sound wave approaches the observer:**

Let's assume that the actual frequency $f$ emitted by the source is 100 Hz.

Given, the frequency changes by 20% as it approaches the observer.

So, the new frequency $f'$ received by the observer is:

$f' = \left(1 + \frac{v_s}{v} \right)f = \left(1 + \frac{0.2v}{v} \right)100 = 1.2 \times 100 = 120$

Now, let's find the velocity of the source and observer when the sound wave approaches the observer.

Given, the frequency changes by 20% as it approaches the observer.

So, we can use the following formula to find the velocity of the source and observer:

$\frac{v_s \pm v}{v} = \pm \frac{f'}{f' - f}$

- When the source is approaching the observer, we use the positive sign in the above formula.
- When the source is receding away from the observer, we use the negative sign in the above formula.

So, when the sound wave approaches the observer, we have:

$\frac{v_s + v}{v} = \frac{120}{120 - 100}$

$\frac{v_s + v}{v} = \frac{6}{5}$

$v_s + v = \frac{6}{5}v$

Similarly, we have:

$\frac{v_o + v}{v} = -\frac{120}{120 - 100}$

$\frac{v_o + v}{v} = -\frac{6}{5}$

$v_o + v = -\frac{6}{5}v$

**When sound wave recedes away from the observer:**

Now, let's find the velocity of the source and observer when the sound wave recedes away from the observer.

Given, the frequency changes by 20% as it recedes away from the observer.

So, we can use the following formula