Question

An A.P consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 39th term.

Answer 1

Hint: Here, we will find the constant difference between the two numbers in series and then we will find the 30th term of A.P. We will be using the formula for the nth term of an A.P, given by ${{a}_{n}}=a+\left( n-1 \right)d$.

Complete step-by-step solution:
It is given in the question that there are 50 terms and all the terms are in A.P. Also, the third term in A.P is 12 and the last term in A.P is 106. And we have to find the 30th term of A.P.
To solve this question, let us first see what an arithmetic progression is. An arithmetic progression is a sequence of numbers such that the difference of any two successive numbers is a constant. For example consider the series: 3, 6, 9, 12, 15. In this series, we have the common difference between any two successive numbers as 3. Hence we know that they are in arithmetic progression.
We know that the general term of any A.P is represented as: ${{a}_{n}}=a+\left( n-1 \right)d.........\left( i \right)$, where, n is the number of terms, d is the common difference and a is the first term.
So, we can express the third term in the general form by substituting n = 3 in equation (i). Hence we get,
${{a}_{3}}=a+\left( 3-1 \right)d.........\left( ii \right)$
We have ${{a}_{3}}=12$. So by substituting the value of ${{a}_{3}}$ in equation (ii) we get,
$12=a+2d..........\left( iii \right)$
Similarly, we have ${{a}_{50}}=106$, so on substituting the value of ${{a}_{50}}$ in equation (i) we get,
$\begin{align}
  & 106=a+\left( 50-1 \right)d \\
 & 106=a+49d...........\left( iv \right) \\
\end{align}$
On subtracting equation (iv) from equation (iii) we get,
\[\begin{align}
  & \dfrac{\begin{align}
  & \text{ }a+2d=12 \\
 & -\text{ }a+49d=106 \\
\end{align}}{0-47d=-94} \\
 & d=\dfrac{94}{47} \\
 & d=2 \\
\end{align}\]
So we get the value of d as 2. Now, substituting the value of d in equation (iii) we get,
$\begin{align}
  & a+2(2)=12 \\
 & a=12-4 \\
 & a=8 \\
\end{align}$
Thus the first term of A.P is 8 and the common difference is 2. Now to find the 30th term of A.P we get,
${{a}_{30}}=a+\left( 30-1 \right)d.........\left( v \right)$
We have the values of a = 8 and d = 2, so on substituting the values of a and d in equation (v) we get,
$\begin{align}
  & {{a}_{30}}=8+\left( 30-1 \right)2 \\
 & {{a}_{30}}=8+29\times 2 \\
 & {{a}_{30}}=8+58 \\
 & {{a}_{30}}=66 \\
\end{align}$
Thus we get the 30th term of A.P as 66.

Note: Students must make sure that they use the correct formula for finding the general term of A.P that is ${{a}_{n}}=a+\left( n-1 \right)d$ and must make a proper substitution of the values of a and d where a is the first term and d is the common difference. The sum of the terms in an A.P is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Also if a, b, c are in A.P, then we can write that a + c = 2b.