An A.P. consists of 50 terms in which the $3^{\text{rd}}$ term is 12 and the last term is 106. Find the $29^{\text{th}}$ term.

(Hint: If $a$ is the first term and $d$ the common difference, then we arrive at the equations $a + 2d = 12$ and $a + 49d = 106$. Solve this pair of linear equations for $a$ and $d$.)                                                                 [Page No. 185]

Let $a =$ first term, $d =$ common difference.

$$3^{\text{rd}} \text{ term} = 12$$

Last term ($50^{\text{th}}$ term) = 106

$$\text{Using: } a_n = a + (n - 1)d$$

$$3^{\text{rd}} \text{ term: } a + 2d = 12$$

$$50^{\text{th}} \text{ term: } a + 49d = 106$$

$$\text{Subtract: } 47d = 94$$

$$d = 2$$

$$\text{Now, } a + 2(2) = 12$$

$$a + 4 = 12$$

$$a = 8$$

$$29^{\text{th}} \text{ term: } a_{29} = 8 + (29 - 1) \times 2$$

$$= 8 + 56$$

$$= 64$$

Therefore, the $29^{\text{th}}$ term is 64.