Question

The 16th term of an AP is 1 more than twice its 8th term. If the 12th term of the AP is 47 then find its nth term.

Answer 1

given,
a + 15d = 1 + 2(a + 7d)
a + 15d = 1+ 2a + 14d
a-d = -1. [1]

a + 11d = 47. [2]
by elimination method , we get d = 4 and a = 3

an= a + (n-1)d
an = 3 + ( n-1 ) 4
an = 3 + 4n -4
an = 4n -1
here is your answer....
hope it helps uu........

Answer 2

$\underline{\mathfrak{Solution : }}$


$\mathsf{Formula \: to \: be \: used : } \\ \\ \boxed{\mathsf{\implies T_{n} \: = \: a \: + \: ( n \: - \: 1 )d }}$


$\underline{\mathsf{Case \: 1 ,}}$


$\mathsf{\implies T_{16} \: = \: 2 \: \times \: T_{8} \: + \: 1} \\ \\ \mathsf{ \implies a \: + \: (16 \: - \: 1)d \: = \: 2[a \: + \: ( 8 \: - \:1)d ] \: + \: 1} \\ \\ \mathsf{ \implies a \: + \: 15d \: = \: 2(a \: + \: 7d) \: + \: 1} \\ \\ \mathsf{ \implies a \: + \: 15d \: = \: 2a \: + \: 14d \: + \: 1} \\ \\ \mathsf{ \implies 15d \: - \: 14d \: = \: 2a \: - \: a \: + \: 1} \\ \\ \mathsf{ \implies d \: = \: a \: + \: 1 \qquad...(1)}$


$\underline{\mathsf{Case \: 2 , }}$

$\mathsf{\implies T_{12} \: = \: 47 } \\ \\ \mathsf{\implies a \: + \: ( 12 \: - \: 1)d \: = \: 47 } \\ \\ \mathsf{\implies a \: + \: 11d \: = \: 47 } \\ \\ \mathsf{\therefore \quad a \: = \: 47 \: - \: 11d \qquad(2) } \\ \\ \mathsf{Plug \:the \: value \: of \: ( 2 ) \: in \: (1)}$


$\mathsf{\implies d \: = \: a \: + \: 1 } \\ \\ \mathsf{\implies d \: = \: 47 \: - \:11d \: + \: 1} \\ \\ \mathsf{\implies d \: + \: 11d \: = \: 47 \: + \: 1} \\ \\ \mathsf{\implies 12d \: = \: 48} \\ \\ \mathsf{\implies d \: = \: \dfrac{48}{12}} \\ \\ \mathsf{\therefore \quad d \: = \: 4 }$

$\mathsf{Plug \: the \: value \: of \: \textbf{d} \: in \: (1),} \\ \\ \mathsf{\implies 4 \: = \: a \: + \: 1} \\ \\ \mathsf{\implies 4 \: - \:1 \: = \: a } \\ \\ \mathsf{\implies 3 \: = \: a } \\ \\ \mathsf{\therefore \quad a \: = \: 3 }$


$\mathsf{Now,} \\ \\ \mathsf{\implies T_{n} \: = \: a \: + \: ( n \: - \: 1)d} \\ \\ \mathsf{Plug \: the \: value \: of \: \textbf{a} \: and \: \textbf{d},}$


$\mathsf{\implies T_{n} \: = \: 3 \: + \:( n \: - \: 1)4} \\ \\ \mathsf{\implies T_{n} \: = \: 3 \: + \: 4n \: - \:4 } \\ \\ \mathsf{\implies T_{n} \: = \: 4n \: - \: 4 \: + \: 3} \\ \\ \mathsf{\therefore \quad T_{n} = \: 4n \: - \: 1 }$