在上个月的第一天，一家制造厂内一台机器的温度被发现为 $90^\circ\text{C}$，主管认为该温度过高。每天，机器产生的废热作为副产品会导致机器温度上升 $2^\circ\text{C}$。为了降低温度，主管决定调整恒温器，并在每天结束时将温度降低 $3\%$。

1. Calculate the temperature at the end of the 1st day.

Let $T_n$ be the temperature at the end of the $n^{\text{th}}$ day.
During the day, the temperature increases by $2^\circ\text{C}$, and at the end of the day it decreases by $3\%$, which corresponds to multiplying by $(1 - 0.03) = 0.97$.
$T_1 = (90 + 2) \times 0.97$
$T_1 = 92 \times 0.97 = 89.24^\circ\text{C}$

2. Calculate the temperature at the end of the 2nd day.

$T_2 = (T_1 + 2) \times 0.97$
$T_2 = (89.24 + 2) \times 0.97$
$T_2 = 91.24 \times 0.97 = 88.5028^\circ\text{C}$

3. Calculate the temperature at the end of the 3rd day and round to show the required value.

$T_3 = (T_2 + 2) \times 0.97$
$T_3 = (88.5028 + 2) \times 0.97$
$T_3 = 90.5028 \times 0.97 = 87.787716^\circ\text{C}$
Rounding to 1 decimal place:
$T_3 \approx 87.8^\circ\text{C} \text{ (shown)}$

1. Express the temperature at the end of day $n$, $u_n$, by establishing a pattern from the first few days.

\text{Let } u_n \text{ be the temperature at the end of the } n\text{-th day.}
\text{Beginning of day 1: } 90^\circ\text{C}
u_1 = (90 + 2)(0.97) = 90(0.97) + 2(0.97)
u_2 = (u_1 + 2)(0.97) = 90(0.97)^2 + 2(0.97)^2 + 2(0.97)
u_3 = (u_2 + 2)(0.97) = 90(0.97)^3 + 2(0.97)^3 + 2(0.97)^2 + 2(0.97)

\text{By pattern recognition, for the } n\text{-th day:}
u_n = 90(0.97)^n + \left[ 2(0.97)^n + 2(0.97)^{n-1} + \dots + 2(0.97) \right]

2. Simplify the expression for $u_n$ using the formula for the sum of a geometric series.

\text{The terms in the square brackets form a geometric progression with first term } a = 2(0.97) = 1.94, \text{ common ratio } r = 0.97, \text{ and } n \text{ terms.}
S_n = \frac{1.94(1 - 0.97^n)}{1 - 0.97} = \frac{1.94}{0.03}(1 - 0.97^n) = \frac{194}{3}(1 - 0.97^n)

\text{Substituting this back into } u_n:
u_n = 90(0.97)^n + \frac{194}{3}(1 - 0.97^n)
u_n = \left(90 - \frac{194}{3}\right)(0.97)^n + \frac{194}{3}
u_n = \frac{76}{3}(0.97)^n + \frac{194}{3}

3. Set up the inequality $u_n < 70$ and solve for $n$ using logarithms.

\text{We want to find the smallest integer } n \text{ such that } u_n < 70:
\frac{76}{3}(0.97)^n + \frac{194}{3} < 70
76(0.97)^n + 194 < 210
76(0.97)^n < 16 \implies 0.97^n < \frac{16}{76} = \frac{4}{19}

\text{Taking the natural logarithm on both sides:}
n \ln(0.97) < \ln\left(\frac{4}{19}\right)

\text{Since } \ln(0.97) < 0 \text{, the inequality sign flips when dividing:}
n > \frac{\ln(4/19)}{\ln(0.97)}
n > 51.155

4. Determine the exact day the temperature criteria is met.

\text{Since } n \text{ must be a whole number of days, the smallest possible integer value is } n = 52.

1. Formulate the general expression for the temperature at the end of $n$ days (building on the pattern from earlier parts).

Let $u_n$ be the temperature at the end of the $n^{\text{th}}$ day. Based on the daily percentage decrease and constant increase, we have:
$u_n = 90(0.97)^n + 2(0.97)^n + 2(0.97)^{n-1} + \dots + 2(0.97)$
The terms after the first form a finite geometric series with first term $a = 2(0.97) = 1.94$ and common ratio $r = 0.97$.
$u_n = 90(0.97)^n + \frac{1.94(1 - 0.97^n)}{1 - 0.97}$

2. Evaluate the limit as $n \to \infty$ to justify whether the temperature drops indefinitely.

As $n \to \infty$, since the common ratio $|0.97| < 1$, the term $0.97^n \to 0$.
Therefore, the sequence converges to a finite limit and the temperature does not drop indefinitely.
$u_n \to 90(0) + \frac{1.94(1 - 0)}{1 - 0.97}$

3. Calculate the exact long term equilibrium temperature.

$\text{Long term temperature} = \frac{1.94}{0.03} = \frac{194}{3} = 64\frac{2}{3}^\circ\text{C}$
Conclusion: No, it will not drop indefinitely. It stabilises at $64\frac{2}{3}^\circ\text{C}$ (or $64.7^\circ\text{C}$ to 3 s.f.).