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

Let a be the first term and d be the common difference of the given A.P.
According to the given question,
$16$th term of the AP $=2\times$ $8^{th}$ term of the AP + 1
i.e., $a_{16}=2a_8+1$
We know that $a_n=a+(n-1)d$
$a+(16-1)d=2[a+(8-1\left)d\right]+5$
$\Rightarrow a+15d=2[a+7d]+5$
$\Rightarrow a+15d=2a+14d+5$
$\Rightarrow d=a+1$--- (1)
Also, ${12}^{th}$ term, $a_{12}=47$
$\Rightarrow a+(12-1)d=47$
$\Rightarrow a+11d=47$
$\Rightarrow a+11(a+1)=47$
$\Rightarrow a+11a+11=47$
$\Rightarrow 12a=36$
$\Rightarrow a=3$
On Putting the value of a in (1), we get $d=3+1=4$
Thus, $n^{th}$ term of the AP, $a_n=a+(n-1)d$
On putting the respective values of a and d, we get
$a_n=3+(n-1)4=3+4n-4=4n-1$
Hence, $n_{th}$ term of the given AP is $4n-1$.