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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,

\[{{a}_{29}}=8+\left( 29-1 \right)2\]

\[\Rightarrow {{a}_{29}}=8+28\times 2\]

\[\Rightarrow {{a}_{29}}=8+56\]

\[\Rightarrow {{a}_{29}}=64\]