Question

Answer the following
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.मुझे जवाब नहीं चाहिए ​

Answer 1

$\mathtt{ n=50, \; T_{3} = 12, \; l = T_{50} = 106}$

$\mathtt{T_{29} = ?}$

$\mathtt{ \;\;\;\;\;\;\;\; T_{3} = 12 }$

$\implies \mathtt{a+2d = 12 \;\;\;\;\; ----(i)}$

$\mathtt{\;\;\;\;\;\;\;\; T_{50} = 106}$

$\implies \mathtt{a+49d = 106 \;\;\;\; ----(ii)}$

$\texttt{Subtracting (i) from (ii)}$

$\mathtt{\;\;\;\;\; a + 49d = 106}$

$\mathtt{\;\;\;\;\; a + \;\;2d = \;\;12}$

___________________

$\texttt{\;\;\;\;\;\;\;\; 47d = 94}$

$\implies \mathtt{ d = \big \frac{94}{47}}$

$\implies \mathtt{ d = 2}$

$\texttt{Putting d=2 in eq (i)}$

$\mathtt{\;\;\;\;\;\;\;\;a+49×2 = 106}$

$\implies \mathtt{a+98 = 106}$

$\implies \mathtt{a = 106-98 = 8}$

$\mathtt{T_{29} = a+28d = 8 + 28×2 = 8+56 = 64}$