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Question
An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term of the A.P.
Solution
Given that,
a3 = 12
a50 = 106
We know that,
an = a + (n − 1)d
a3 = a + (3 − 1)d
12 = a + 2d ...(i)
Similarly, a50 = 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(2)
a = 12 − 4
a = 8
a29 = a + (29 − 1)d
a29 = 8 + (28)2
a29 = 8 + 56
a29 = 64
Therefore, 29th term is 64.
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