To solve the problem, let's first recall the formula for the $n$th term of an arithmetic sequence:

$a_n = a_1 + (n - 1) \cdot d$

Where:

• $a_n$ is the $n$th term,
• $a_1$ is the first term,
• $d$ is the common difference,
• $n$ is the term number.

Given:

• The fourth term $a_4 = 10$,
• The tenth term $a_{10} = 40$.

Step 1: Set up the equations

For the fourth term ($n = 4$): $a_4 = a_1 + 3d = 10$

For the tenth term ($n = 10$): $a_{10} = a_1 + 9d = 40$

Step 2: Subtract the first equation from the second

$(a_1 + 9d) - (a_1 + 3d) = 40 - 10$ $6d = 30$ $d = 5$

Step 3: Solve for $a_1$

Substitute $d = 5$ into the first equation: $a_1 + 3 \times 5 = 10$ $a_1 + 15 = 10$ $a_1 = -5$

Step 4: Find the 25th term

Now, use the formula for the $n$th term to find the 25th term ($n = 25$): $a_{25} = a_1 + 24d$ $a_{25} = -5 + 24 \times 5$ $a_{25} = -5 + 120$ $a_{25} = 115$

So, the 25th term of the sequence is 115.

Would you like more details or have any questions about this?

Here are 5 related questions for further practice:

1. What is the first term of the sequence if the 3rd term is 12 and the 7th term is 36?
2. How many terms are there in an arithmetic sequence if the first term is 4, the last term is 100, and the common difference is 3?
3. What is the sum of the first 20 terms of an arithmetic sequence where the first term is 7 and the common difference is 2?
4. Find the 50th term of an arithmetic sequence where the 5th term is 20 and the common difference is 4.
5. Determine the common difference of an arithmetic sequence if the first term is 8 and the 10th term is 50.

Tip: In arithmetic sequences, the difference between any two consecutive terms is always constant. This property is key to solving many problems involving such sequences.