Question

An A-P consists of 50 terms of which 3rd term is 12 and the the last term is 106 Its 29th term is

• A. 58
• B. 60
• C. 61
• D. 64

Answer 1

The correct option is D 64

$t_3=12$ and $t_{50}=106$
$\Rightarrow a=2d=12$ ....(i)
and $a+49d=106$ ...(ii)
(i) and (ii) gives a = 8, d = 2
$\therefore t_{29}=a+28d=8+56=64$