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Question: The \({{16}^{th}}\) term of an AP is 1 more than twice its \({{8}^{th}}\) term. If the \({{12}^{th}}...

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

Explanation

Solution

With the help of the formula of nthn^{th} term an=a+(n−1)d{{a}_{n}}=a+\left( n-1 \right)d, we will find the 16th{{16}^{th}} , 8th{{8}^{th}} and 12th{{12}^{th}} term. Then according to question, we will substitute these in equations to find the nthn^{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 nthn^{th} term an{{a}_{n}} of AP is calculated by the formula
an=a+(n−1)d …(i){{a}_{n}}=a+\left( n-1 \right)d\text{ }\ldots \left( i \right)
Here, the 16th{{16}^{th}} term is given by a16=a+(16−1)d=a+15d{{a}_{16}}=a+\left( 16-1 \right)d=a+15d
8th{{8}^{th}} term is given by a8=a+(8−1)d=a+7d{{a}_{8}}=a+\left( 8-1 \right)d=a+7d
12th{{12}^{th}} term is given by a12=a+(12−1)d=a+11d{{a}_{12}}=a+\left( 12-1 \right)d=a+11d
According to question,
a16=2a8+1{{a}_{16}}=2{{a}_{8}}+1
Putting the values in above equation, we get

& \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{aligned}$$ Also, it is given in question that $\begin{aligned} & \text{ }{{a}_{12}}=47 \\\ & \Rightarrow a+11d=47\text{ }\ldots \left( iii \right) \\\ \end{aligned}$ Solving equations (ii) and (iii) by substitution method, we get $\begin{aligned} & \text{ }a+11d=47 \\\ & \Rightarrow d - 1+11d=47 \\\ & \Rightarrow 12d=48 \\\ & \Rightarrow d=4 \\\ \end{aligned}$ Putting value of d in equation (ii), we get $\begin{aligned} & a=d-1 \\\ & \Rightarrow a=4-1\\\ & \Rightarrow a= 3\\\ \end{aligned}$ Putting values of a and d in equation (i) to find the value of $n^{th}$ term, we get $\begin{aligned} & {{a}_{n}}=3+\left( n-1 \right)4 \\\ & \Rightarrow {{a}_{n}}=3+4n-4 \\\ & \Rightarrow {{a}_{n}}=4n-1 \\\ \end{aligned}$ **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.