Question 4

We need to find the 29th term of the arithmetic progression.

Step 1 — Find 'a' and 'd'

Let's define the first term as $a$. Let's define the common difference as $d$. The $n^{\text{th}}$ term of an AP is $a_n = a + (n-1)d$. We are given that the $3^{\text{rd}}$ term is 12.

$a + (3-1)d = 12$

$a + 2d = 12 \quad \text{(Equation 1)}$

The AP has 50 terms. The last term is 106. So, the $50^{\text{th}}$ term is 106.

$a + (50-1)d = 106$

$a + 49d = 106 \quad \text{(Equation 2)}$

Now, let's solve these two equations. We will subtract Equation 1 from Equation 2.

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

$47d = 94$

$\boxed{d = 2}$

Now, let's substitute $d = \mathbf{2}$ into Equation 1.

$a + 2(\mathbf{2}) = 12$

$a + 4 = 12$

$a = 12 - 4$

$\boxed{a = 8}$

Step 2 — Find the 29th term

We need to find the $29^{\text{th}}$ term. We use the formula $a_n = a + (n-1)d$. We know $a = \mathbf{8}$ and $d = \mathbf{2}$.

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

$a_{29} = \mathbf{8} + (28)\mathbf{2}$

$a_{29} = 8 + 56$

$\boxed{a_{29} = 64}$

Answer

The 29th term is 64.