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

An AP consists of 50 terms and the $$50^{th}$$ term is equal to 106 and $$a_3 = 12$$

Using formula $$a_n = a + (n - 1)d$$ , to find nth term of arithmetic progression,

$$\Rightarrow a_{50}$$= a + (50 - 1)d, and
$$a_3$$= a + (3 - 1)d
$$\Rightarrow$$ 106 = a + 49d and,
12 = a + 2d

These are equations consisting of two variables.

Using equation 106 = a + 49d,
we get a = 106 - 49d

Putting value of a in the equation 12 = a + 2d,

$$\Rightarrow$$ 12 = 106 - 49d + 2d
$$\Rightarrow$$ 47d = 94
$$\Rightarrow$$ d = 2

Putting value of d in the equation,
a = 106 - 49d,

a = 106 – 49 (2)
= 106 – 98 = 8

Therefore, First term = a = 8 and
Common difference = d = 2

To find $$29^{th}$$ term,
we use formula $$a_n = a + (n - 1)d$$ which is used to find nth term of arithmetic progression,

$$a_{29}$$
= 8 + (29 - 1) 2
= 8 + 56 = 64

Therefore, 29th term of AP is equal to 64.