Which term of the A.P $\;$ $5, \; 12, \; 19, \; 26, \; 33, \; \cdots$ $\;$ will be $35$ more than its $12^{th}$ term?
Given A.P : $\;$ $5, \; 12, \; 19, \; 26, \; 33, \; \cdots$
First term of A.P $= a = 5$
Common difference of A.P $= d = 7$
Let $n^{th}$ term of A.P be $35$ more than its $12^{th}$ term.
$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right)d = 5 + \left(n - 1\right) \times 7$
$12^{th}$ term of A.P $= t_{12} = 5 + \left(12 - 1\right) \times 7 = 82$
As per question,
$5 + \left(n - 1\right) \times 7 = 82 + 35$
i.e. $\;$ $5 + 7n - 7 = 117$
i.e. $\;$ $7n = 105$ $\implies$ $n = 15$
$\therefore \;$ The $15^{th}$ term of the given A.P will be $35$ more than its $12^{th}$ term.