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Question:

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

Solution:

Given,a3=12,a50=106an=a+(n−1;)da3=a+(3−1;)d12=a+2d... (i)a50=a+(50−1;)d106=a+49d... (ii)On subtracting (i) from (ii), we get94=47d∴d=2From equation (i), we get12=a+2(2)⇒a=12−4=8Nowa29=a+(29−1;)d=8+(28)2=8+56=64Therefore,29thterm is64.