# An AP consists of

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 50th term is 106, substituting this in general term, we get

a50 = a + (50 - 1)d = 106

a + 49d = 106…………(iii)

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

a + 49d = 106

- (a + 2d = 12)

47 d = 94

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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