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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.

Last updated date: 29th Feb 2024
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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.

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.