The 21st and 33rd terms of an arithmetic progression are 91 and 145 r…

The correct answer is Option (1) → 127

Let's solve this step by step.

We are given:

• $a_{21} = 91$
• $a_{33} = 145$

Step 1: Use the formula for the n-th term of an AP:

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

where a = first term, = common difference.

Step 2: Write equations for the given terms:

$a + 20d = 91 \quad \text{(1)}$

$a + 32d = 145 \quad \text{(2)}$

Step 3: Subtract (1) from (2):

$(a + 32d) - (a + 20d) = 145 – 91$

$12d = 54 \Rightarrow d = 4.5$

Step 4: Find a:

$a + 20(4.5) = 91$

$a + 90 = 91 \Rightarrow a = 1$

Step 5: Find the 29th term:

$a_{29} = a + 28d = 1 + 28(4.5) = 1 + 126 = 127$