Q. 244.8( 23 Votes )

An AP consists of

Answer :

Given AP has 50 terms,


So general term can be written as,


an = a + (n – 1) d………..(i)


Now given the AP’s 3rd term is 12, substituting this in general term, we get


a3 = a + (3 – 1)d = 12


a + 2d = 12…………(ii)


It is also given the last term is 106, as we know the AP has 50 terms,

so the 50th term is 106,


substituting this in general term, we get,


a50 = 106

⇒  a + (50 – 1)d = 106

a + 49d = 106…………(iii)


Now subtracting equation (ii) from equation (iii), we get



d = 2


Now substituting the value of d in equation (ii), we get


a + 2d = 12


a + 2(2) = 12


a + 4 = 12


a = 12 – 4


a = 8


So, the given AP has first term, a = 8 and the difference, d = 2,


so the 29th term will have n = 29,


substituting these values in equation (i), we get


a29 = a + (29 – 1)d


a29 = 8 + (28)2


a29 = 8 + 56


a29 = 64


Hence the 29th term of the given AP is 64.

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