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., $a_{16}=2\times a_8+1$
$\Rightarrow a+(16-1)d=2[a+(8-1\left)d\right]+1$ [$\because$ $a_n=a+(n-1)d]$
$\Rightarrow a+15d=2[a+7d]+1$
$\Rightarrow a+15d=2a+14d+1$
$\Rightarrow d=a+1....\left(1\right)$
Also, ${12}^{th}term,a_{12}=47$ [Given]
$\Rightarrow a+(12-1)d=47$
$\Rightarrow a+11d=47$
$\Rightarrow a+11(a+1)=47\left[Using\right(1\left)\right]$
$\Rightarrow a+11a+11=47$
$\Rightarrow 12a=36$
$\Rightarrow a=3$
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 $4n-1.$