Question

The ${{16}^{th}}$ term of an AP is 1 more than twice its ${{8}^{th}}$ term. If the ${{12}^{th}}$term of the AP is 47, then find its $n^{th}$ term.

Answer 1

Hint: With the help of the formula of $n^{th}$ term ${{a}_{n}}=a+\left( n-1 \right)d$, we will find the ${{16}^{th}}$ , ${{8}^{th}}$ and ${{12}^{th}}$ term. Then according to question, we will substitute these in equations to find the $n^{th}$ term.

Complete step by step answer:
We know, in an AP, let a be its first term, n be the number of terms and d be the common difference. Then the $n^{th}$ term ${{a}_{n}}$ of AP is calculated by the formula
${{a}_{n}}=a+\left( n-1 \right)d\text{ }\ldots \left( i \right)$
Here, the ${{16}^{th}}$ term is given by ${{a}_{16}}=a+\left( 16-1 \right)d=a+15d$
${{8}^{th}}$ term is given by ${{a}_{8}}=a+\left( 8-1 \right)d=a+7d$
${{12}^{th}}$ term is given by ${{a}_{12}}=a+\left( 12-1 \right)d=a+11d$
According to question,
${{a}_{16}}=2{{a}_{8}}+1$
Putting the values in above equation, we get
\[\begin{align}
  & \text{ }a+15d=2\left( a+7d \right)+1 \\
 & \Rightarrow a+15d=2a+14d+1 \\
 & \Rightarrow -a+d=1 \\
 & \Rightarrow a=d - 1\text{ }\ldots \left( ii \right) \\
\end{align}\]
Also, it is given in question that
$\begin{align}
  & \text{ }{{a}_{12}}=47 \\
 & \Rightarrow a+11d=47\text{ }\ldots \left( iii \right) \\
\end{align}$
Solving equations (ii) and (iii) by substitution method, we get
$\begin{align}
  & \text{ }a+11d=47 \\
 & \Rightarrow d - 1+11d=47 \\
 & \Rightarrow 12d=48 \\
 & \Rightarrow d=4 \\
\end{align}$
Putting value of d in equation (ii), we get
$\begin{align}
  & a=d-1 \\
 & \Rightarrow a=4-1\\
 & \Rightarrow a= 3\\
\end{align}$
Putting values of a and d in equation (i) to find the value of $n^{th}$ term, we get
$\begin{align}
  & {{a}_{n}}=3+\left( n-1 \right)4 \\
 & \Rightarrow {{a}_{n}}=3+4n-4 \\
 & \Rightarrow {{a}_{n}}=4n-1 \\
\end{align}$

Hence, the $n^{th}$ term is given by ${{a}_{n}}=4n-1$

Note: We can solve the two linear equations by elimination method and cross-multiplication method. But the coefficients of a and d were small and for ‘a’ it was 1 in both cases. So, other methods would have led to unnecessary complex calculations which could lead to an error. Hence, we chose a substitution method.