A triode has a plate resistance of 8 k and a transconductance of...

Given:
Plate resistance $r_p = 8 \text{ k}\Omega = 8 \times 10^3 \Omega$
Transconductance $g_m = 2.5 \text{ millimho} = 2.5 \times 10^{-3} \text{ mho}$
Change in plate voltage $\Delta V_p = 48 \text{ V}$

The amplification factor $\mu$ is given by the product of plate resistance and transconductance:
$\mu = r_p \times g_m$
$\mu = (8 \times 10^3) \times (2.5 \times 10^{-3}) = 20$

For a constant plate current ($\Delta I_p = 0$), the relation between the change in plate voltage and the change in grid voltage is:
$\mu = -\dfrac{\Delta V_p}{\Delta V_g}$

Substituting the values:
$20 = -\dfrac{48}{\Delta V_g}$
$\Delta V_g = -\dfrac{48}{20} = -2.4 \text{ V}$

The negative sign indicates that the grid voltage must be reduced. The required reduction is $2.4 \text{ V}$.