Question

The seventh term of an Arithmetic progression is four times its second term and twelfth term is 2 more than three times of its fourth term. Find the progression.

Answer 1

Hint: We will start by using the fact that the ${{n}^{th}}$ term of an AP with first term a and common difference d is $a+\left( n-1 \right)d$. Then using this fact we will find the second, fourth and twelfth term of the AP and use the data given in the question to form equations and solve them to find the series.

Complete step-by-step answer:
Now, we will consider the first term of the progression as a and common difference as d. Also, we will consider that we have the AP as $a,a+d,a+2d......$
Now, we know the for an AP with first term a and common difference d the ${{n}^{th}}$ term is $a+\left( n-1 \right)d$.
Now, we have seventh term of AP as,
$\begin{align}
  & a+\left( 7-1 \right)d \\
 & \Rightarrow a+6d..........\left( 1 \right) \\
\end{align}$
Also, similarly we have second term of the AP as,
$\begin{align}
  & a+\left( 2-1 \right)d \\
 & \Rightarrow a+d..........\left( 2 \right) \\
\end{align}$
Now, we have the fourth term of AP as,
$\begin{align}
  & a+\left( 4-1 \right)d \\
 & \Rightarrow a+3d..........\left( 3 \right) \\
\end{align}$
Now, we have been given the question that the seventh term is 4 times the second term. So, we have from (1) and (2).
$\begin{align}
  & a+6d=4\left( a+d \right) \\
 & a+6d=4a+4d \\
 & 6d-4d=4a-a \\
 & 2d=3a.........\left( 4 \right) \\
\end{align}$
Now, we have been given that twelfth term is 2 more than three times of its fourth term. So, we have,
$\begin{align}
  & {{12}^{th}}term=2+3\times {{4}^{th}}term \\
 & a+11d=2+3\left( a+3d \right) \\
 & a+11d=2+3a+9d \\
 & 11d-9d=2+3a-a \\
 & 2d=2+2a \\
 & d=1+a.........\left( 5 \right) \\
\end{align}$
Now, we will substitute the value of d from (5) in (4). So, we have,
$\begin{align}
  & 2\left( 1+a \right)=3a \\
 & 2+2a=3a \\
 & 2=3a-2a \\
 & 2=a \\
\end{align}$
Now, substituting $a=2$ in (5) we have,
$\begin{align}
  & d=1+2 \\
 & d=3 \\
\end{align}$
So, we have the progression as
$\begin{align}
  & a,a+d,a+2d...... \\
 & \Rightarrow 2,2+3,2+2\times 3,...... \\
 & \Rightarrow 2,5,8,11,....... \\
\end{align}$


Note: It is important to note that we have used a fact that an arithmetic progression is of type $a,a+d,a+2d......$ where a is the first term and d is the common difference. Also, we have used the formula for ${{n}^{th}}$ term of AP i.e. ${{n}^{th}}term=a+\left( n-1 \right)d$ to find the second, fourth, seventh and twelfth term of the series. The probable mistake that we can make is misinterpreting the given data and forming the wrong equations. We might wrongly form the second equation as \[{{12}^{th}}term+2=3\times {{4}^{th}}term\] and then get the wrong answer.