The temperature of a metal strip having coefficient of linear...

Let the length of the metal strip at temperature $T_1$ be $L_1$.

When the temperature is increased from $T_1$ to $T_2$, the new length $L_2$ is given by:

$L_2 = L_1(1 + \alpha \Delta T)$

where $\Delta T = T_2 - T_1$.

The increase in length is:

$\Delta L_1 = L_2 - L_1 = L_1 \alpha \Delta T$

When the temperature is further increased from $T_2$ to $T_3$, the new length $L_3$ is:

$L_3 = L_2(1 + \alpha \Delta T')$

where $\Delta T' = T_3 - T_2$.

Given $T_3 + T_1 = 2T_2$, we have $T_3 - T_2 = T_2 - T_1 = \Delta T$.

Thus, $\Delta T' = \Delta T$.

The increase in length for this second temperature change is:

$\Delta L_2 = L_3 - L_2 = L_2 \alpha \Delta T$

Substituting $L_2 = L_1(1 + \alpha \Delta T)$ into the equation for $\Delta L_2$:

$\Delta L_2 = L_1(1 + \alpha \Delta T) \alpha \Delta T$

$\Delta L_2 = (L_1 \alpha \Delta T)(1 + \alpha \Delta T)$

Since $\Delta L_1 = L_1 \alpha \Delta T$, substituting this gives:

$\Delta L_2 = \Delta L_1(1 + \alpha \Delta T)$

Answer: $\Delta L_1[1+\alpha\Delta T]$