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, a3 = 12 and a50 = 106

a3 = a + 2d = 12

a50 = a + 49d = 106

Subtracting 3rd term from 50th term, we get;

a + 49d – a – 2d = 106 – 12

Or, 47d = 94

Or, d = 2

Substituting the value of d in 12th term, we get;

a + 2 x 2 = 12

Or, a + 4 = 12

Or, a = 8

Now, 29th term can be calculated as follows:

a29 = a + 28d

= 8 + 28 x 2

= 8 + 56 = 64

Answer 2

$l = 106\\ term(3) = 12\\ \\ = > a + (n - 1)d = 12 \\ = > a + 2d = 12 \: \: \: \: \: ....(1)\\ \\ \\ and\\ \\ \\ l = a + (n - 1)d = 106\\ \\ = > a + 49d = 106 \: \: ....(2)\\ \\ \\ now \\ \\ (2) - (1) \\ we \: get\\ \\ \\ 47d = 94 \\ = > \fbox{d = 2}\\ \\ \\ now\\ \\ \: put \: \fbox{d = 2} \: in \: (1)\\ \\ = > a \: + 2d = 12\\ \\ = > a + 4 = 12 \\ = > \fbox{a = 8} \\ \\ \\ now\\ \\ \\ a \: = 8\\ \\ d = 2\\ \\ \\ so \\ \\ term(29) = a + (29 - 1)d \\ = 8 + 28(2)\\ \\ = 64\\ \\ \\ = > \fbox{term(29) = 64}$

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