8. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Given that, 3rd term, a3 = 12 50th term, a50 = 106 We know that, an = a+(n−1)d a3 = a+(3−1)d 12 = a+2d ……………………………. (i) In the same way, a50 = a+(50−1)d 106 = a+49d …………………………. (ii) On subtracting equation (i) from (ii), we get 94 = 47d d = 2 = common difference From equation (i), we can write now, 12 = a+2(2) a = 12−4 = 8 a29 = a+(29−1) d a29 = 8+(28)2 a29 = 8+56 = 64 Therefore, 29th term is 64.