$\triangle$ Let

$$\text { GCD }(2 a+5,3 a+4)=d \text {. }$$

Therefore,

$$(2 a+5): d, \quad(3 a+4): d$$

Now multiply the number $2 a+5$ by 3, and the number $3 a+4$ by 2. Then

$$(6 a+15): d, \quad(6 a+8): d$$

Subtract the obtained numbers $6 a+15$ and $6 a+8$:

$$7: d \text {. }$$

Hence, $d=1$ or $d=7$. Therefore, if the fraction is reducible, it is only by 7.

Find all natural $a$ for which the fraction is reducible by 7.

The first such value of $a$ is $a=1$, the next is $a=8$, then $a=15,22$, and so on.

The general form of such $a$ is

$$a=1+7 k \quad(k=0,1,2, \ldots)$$

Answer: by 7 when $a=7 k+1(k=0,1,2, \ldots)$.