If the third and ninth terms of an A.P. are 1 and 19 respectively, then the 23rd term will be:

61

The correct answer is Option (2) → 61

Step 1: Let the first term = a and common difference = d

The n-th term of an AP:

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

Step 2: Use given terms

• 3rd term: $T_3 = a + 2d = 1$
• 9th term: $T_9 = a + 8d = 19$

Step 3: Solve for d

$T_9 - T_3 = (a+8d) - (a+2d) = 6d = 19 - 1 = 18$

$d = 3$

Step 4: Find a

$a+2d=1  ⟹  a+6=1  ⟹  a=−5$

Step 5: Find 23rd term

$T_{23} = a + 22d = -5 + 22 \cdot 3 = -5 + 66 = 61$