Question

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

Answer 1

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,

& a+49d=106 \\\ & a+2d=12 \\\ & \underline{-\text{ }-\text{ }-} \\\ & 47d=94 \\\ \end{aligned}$$ Dividing both the sides by 47, we get, $$\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.