Question

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

Answer 1

Given that, $a_3 = 12$ and $a_{50}= 106$
We know that, $a_n = a + (n − 1) d$
$a_3 = a + (3 − 1) d$
$12 = a + 2d$ …..(i)
Similarly,
$a_{50} = a + (50 − 1)d$
$106 = a + 49d$ ……(ii)
On subtracting (i) from (ii), we obtain
$94 = 47d$
$d = 2$
From equation (i), we obtain
$12 = a + 2 \times 2$
$a = 12 − 4$
$a= 8$
$a_{29} = a + (29 − 1)d$
$a_{29} = 8 + 28\times 2$
$a_{29} = 8 + 56 = 64$

Therefore, 29th term is 64.