Question

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the ${29}^{th}$ term.

Answer 1

Hint: We will be using the concepts of series and progression specifically Arithmetic Progression to solve the problem. We will be making an equation with the given condition and then solve it for the answer.

Complete step-by-step solution -
Now, we have been given that an AP consists of 50 terms and the $3^{rd}$ term of the series is 12 and the last term is 106. We have to find the ${29}^{th}$ term. So, let us take the first term of Arithmetic Progression be a, common difference be d. Therefore,
A.P=a, a+d, …………………………………….. a+4d.
Now, we know that the nth term of an A.P is $a+\left( n-1 \right)d$. Therefore, the last term of the AP ${50}^{th}$ term is $a+\left( 50-1 \right)d$ and this has been given equal to 106.
$a+49d=106$ ………………… (1)
Also the $3^{rd}$ term is equal to 12. Therefore,
$a+2d=12$ …………………… (2)
Now, we will solve (1) and (2) for the values of a and d. We will subtract (2) form (1).
$\begin{align}
  & a+49d-a-2d=106-12 \\
 & 47d=94 \\
 & d=\dfrac{94}{47} \\
 & d=2 \\
\end{align}$
Now, we will substitute this value in (2) so that
$\begin{align}
  & a=12-2\times 2 \\
 & a=12-4 \\
 & a=8 \\
\end{align}$
Now, we have to find the 29th term. Therefore,
${29}^{th}$ term $=a+\left( 29-1 \right)d$
$=a+28d$
Now, we will substitute a=8 and d=2 so that
$\begin{align}
  & {{t}_{29}}=8+28\times 2 \\
 & =8+56 \\
 & =64 \\
\end{align}$
The ${29}^{th}$ term is 64.

Note: To solve these types of questions one should first convert all the given conditions into the equations and then solve them to get the answer. Also it is important to remember that the nth term of AP is ${{t}_{n}}=a+\left( n-1 \right)d$ .