Question

An AP consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th 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, 29^th term is 64.