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An AP consists of 50 terms of which the third term is 12 and the last term is 106. Find the \[{{29}^{th}}\] term.

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Hint: In the question, we are given the third term \[{{a}_{3}}\] and the last term \[{{a}_{50}}.\] We will use the formula \[{{a}_{n}}=a+\left( n-1 \right)d\] to find the value of a and d. Once we get the values of a and d, we will use the same formula to find the \[{{29}^{th}}\] term by putting n = 29.

Complete step-by-step answer:
We are given an AP whose third term is 12, that is \[{{a}_{3}}=12\] and the last term is 106, i.e. \[{{a}_{50}}=106.\]
Now, we know that the general term in an AP is given as
\[{{a}_{n}}=a+\left( n-1 \right)d\]
where a is the first term, d is the difference and n is the number of terms.
We have the third term as 12,
\[{{a}_{3}}=12\]
Therefore, using this value in the general formula, we get,
\[{{a}_{3}}=a+\left( 3-1 \right)d\]
\[\Rightarrow 12=a+2d.....\left( i \right)\]
Similarly, we have the last term as 106.
\[{{a}_{50}}=106\]
\[\Rightarrow {{a}_{50}}=a+\left( 50-1 \right)d\]
\[\Rightarrow 106=a+49d.....\left( ii \right)\]
Now, we will solve the for a and d using equation (i) and (ii).
Subtraction equation (ii) from (i), we get,
\[\begin{align}
  & a+49d=106 \\
 & a+2d=12 \\
 & \underline{-\text{ }-\text{ }-} \\
 & 47d=94 \\
\end{align}\]
Dividing both the sides by 47, we get,
\[\Rightarrow d=\dfrac{94}{47}=2\]
Therefore, we get the common difference, d = 2.
Now, putting the value, d = 2 in equation (i), we get,
\[a+2\times 2=12\]
\[\Rightarrow a+4=12\]
\[\Rightarrow a=8\]
Therefore, we get our first term as 8.
Now, we will find the \[{{29}^{th}}\] term. We know that,
\[{{a}_{n}}=a+\left( n-1 \right)d\]
For, \[{{a}_{29}},n=29.\]
Also, we have, a = 8 and d = 2. Therefore, we get,
\[{{a}_{29}}=8+\left( 29-1 \right)2\]
\[\Rightarrow {{a}_{29}}=8+28\times 2\]
\[\Rightarrow {{a}_{29}}=8+56\]
\[\Rightarrow {{a}_{29}}=64\]
So, we get the \[{{29}^{th}}\] term as 64.

Note:While solving for the third term and the last term, students need to remember that the third term is written as \[{{a}_{3}}=a+2d,\] writing \[{{a}_{3}}=a+3d\] will lead to a wrong solution. Also, students have to keep in mind that when subtracting two equations, you need to change the sign of the second equation which is being subtracted.