The temperature at 12 noon was\[{{10}^{0}}C\]above zero. If it decreases at a rate of\[{{2}^{0}}C\]per hour until midnight, at what time would the temperature be\[{{8}^{0}}C\]below zero? What would be the temperature at midnight?
(a) \[{{14}^{0}}C\]
(b) \[-{{14}^{0}}C\]
(c) \[{{16}^{0}}C\]
(d) \[-{{16}^{0}}C\]

Hint: We solve this problem by considering the arithmetic progression having the first term as\[{{10}^{0}}C\]and the common difference as\[-{{2}^{0}}C\]because the temperature is decreasing. The general representation of arithmetic progression is given as
\[a,a+d,a+2d,..........\]
Here, the formula of\[{{n}^{th}}\]is given as
\[{{T}_{n}}=a+\left( n-1 \right)d\]
By using this formula we calculate the required result.

Complete step-by-step answer:
We are given that the initial temperature is\[{{10}^{0}}C\]
We are given that the temperature decreases at a rate of\[{{2}^{0}}C\]per hour.
Let us convert this information into arithmetic progression taking\[{{10}^{0}}C\]as first term and\[-{{2}^{0}}C\]as common difference as follows
\[10,\left( 10+\left( -2 \right) \right),\left( 10+2\left( -2 \right) \right),.........\]
Now, let us find the time at which the temperature is\[{{8}^{0}}C\]below zero
Here, the temperature\[-{{8}^{0}}C\]is the\[{{n}^{th}}\]term.
We know that the general representation of arithmetic progression is given as
\[a,a+d,a+2d,..........\]
Here, the formula of\[{{n}^{th}}\]is given as
\[{{T}_{n}}=a+\left( n-1 \right)d\]
By substituting the required values in above formula we get
\[\begin{align}
  & \Rightarrow -8=10+\left( n-1 \right)\left( -2 \right) \\
 & \Rightarrow \left( n-1 \right)\left( -2 \right)=-18 \\
 & \Rightarrow n-1=9 \\
 & \Rightarrow n=10 \\
\end{align}\]
So, the temperature is\[{{8}^{0}}C\]below zero will occur at\[{{10}^{th}}\]hour from start, so that the time will be 9pm because we considered that the first term as 12 noon.
Therefore at 9pm the temperature is\[{{8}^{0}}C\]below zero.
Now, we are asked to find the temperature at midnight.
Here, we can see that the initial time is taken as 12 noon, so the midnight will come at\[{{13}^{th}}\]term in arithmetic progression because we considered the first term as 12 noon.
So, let us find the temperature at midnight by taking the\[{{n}^{th}}\]term in arithmetic progression as\[{{13}^{th}}\]term.
We know that the general representation of arithmetic progression is given as
\[a,a+d,a+2d,..........\]
Here, the formula of\[{{n}^{th}}\]is given as
\[{{T}_{n}}=a+\left( n-1 \right)d\]
By using this formula we calculate the required result.
Now let us find\[{{13}^{th}}\]term.
By substituting the required values in the formula we get
\[\begin{align}
  & \Rightarrow {{T}_{13}}=10+\left( 13-1 \right)\left( -2 \right) \\
 & \Rightarrow {{T}_{13}}=10+\left( 12 \right)\left( -2 \right) \\
 & \Rightarrow {{T}_{13}}=10-24 \\
 & \Rightarrow {{T}_{13}}=-14 \\
\end{align}\]
Therefore, the temperature at the midnight is\[-{{14}^{0}}C\]

So, the correct answer is “Option (b)”.

Note: Students will make mistakes in taking the number of terms. Here, we have taken the arithmetic progression as
\[10,\left( 10+\left( -2 \right) \right),\left( 10+2\left( -2 \right) \right),.........\]
Here, the first term is\[{{10}^{0}}C\], which is the temperature at 12 noon. Since this is the first term then 12 midnight will occur at\[{{13}^{th}}\]term. But students will make mistakes in taking as\[{{12}^{th}}\]term considering the time, which results in the wrong result. So, the\[{{n}^{th}}\]term depends on the first term. This part needs to be taken care of.