If x is subtracted from each of  23, 39, 32 and 56, the numbers so obtained, in this order, are in proportion. What is the mean proportional between (x + 4) and (3x + 1) ?

As per the question,

$$\Rightarrow \dfrac{23-x}{39-x}=\dfrac{32-x}{56-x}$$

$$\Rightarrow (23-x)(56-x)=(32-x)(39-x)$$

$$\Rightarrow 23\times 56-23x-56x+x^2=32\times 39-32x-39x+x^2$$

$$\Rightarrow 1288-79x+x^2=1248-71x+x^2$$

$$\Rightarrow 79x-71x=1288-1248$$

$$\Rightarrow 8x=40$$

$$\Rightarrow x=5$$

Now, substituting the values in $$=\sqrt{(x+4)(3x+1)}=\sqrt{(5+4)(3\times 5+1)}=\sqrt{9\times 16}=3\times 4=12$$

Hence, the required value=12