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

Given:

An A.P. consists of 50 terms. The 3rd and the last terms are 12 and 106 respectively.

To do:

We have to find the 29th term.

Solution:

Let $a$ be the first term and $d$ be the common difference.

Number of terms $n=50$

3rd term $a_3=a+2d=12$........(i)

Last term $a_n=a+(n-1)d$

Therefore,

$a_{50}=a+(50-1)d=106$

$106=a+49d$.....(ii)

Subtracting (i) from (ii), we get,

$a+49d-a-2d=106-12$

$47d=94$

$d=\frac{94}{47}$

$d=2$

This implies,

$a+2(2)=12$

$a=12-4=8$

29th term $a_{29}=a+(29-1)d$

$=8+28(2)$

$=8+56$

$=64$

The 29th term is 64.

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