The 4th term of an A.P. is 13 while the 10th term is 31. Find the 24th term.

1. Analyze the question

We are given two specific terms of an Arithmetic Progression (A.P.): the 4th term ($T_4 = 13$) and the 10th term ($T_{10} = 31$). We need to find the 24th term ($T_{24}$).

2. Explain the concept

In an Arithmetic Progression, the $n^{th}$ term is calculated using the formula:

Where:

• $a$ is the first term.
• $d$ is the common difference.
• $n$ is the position of the term.

3. Show the solution

Step 1: Set up equations based on the given information. For the 4th term ($n=4$): $a + (4 - 1)d = 13$ $a + 3d = 13$ --- (Equation 1)

For the 10th term ($n=10$): $a + (10 - 1)d = 31$ $a + 9d = 31$ --- (Equation 2)

Step 2: Solve for $d$ by subtracting Equation 1 from Equation 2. $(a + 9d) - (a + 3d) = 31 - 13$ $6d = 18$ $d = \frac{18}{6}$ $d = 3$

Step 3: Solve for $a$ using Equation 1. $a + 3(3) = 13$ $a + 9 = 13$ $a = 13 - 9$ $a = 4$

Step 4: Find the 24th term ($T_{24}$). $T_{24} = a + (24 - 1)d$ $T_{24} = 4 + (23 \times 3)$ $T_{24} = 4 + 69$ $T_{24} = 73$

4. Identify the answer

Comparing our result to the options provided: A. 89 B. 75 C. 73 D. 69

The correct option is C.

5. Tip

In A.P. problems, if you are given two terms, you can find the common difference ($d$) quickly using the slope formula: $d = \frac{T_n - T_m}{n - m}$. Here: $d = \frac{31 - 13}{10 - 4} = \frac{18}{6} = 3$. This saves time during the actual JAMB exam!