In an arithmetic progression, the fourth term is 15 and the fifteenth term is −29. Based on this information, what is the value of the tenth term?

Introduction / Context:
This question requires you to work with the general term of an arithmetic progression and use information about two specific terms to determine the first term and common difference. Once these parameters are known, you can find any other term, including the tenth term. This is a standard type of arithmetic progression problem used to test algebraic manipulation skills.

Given Data / Assumptions:

• Fourth term T4 is 15.
• Fifteenth term T15 is −29.
• The sequence is an arithmetic progression.
• We must find the value of the tenth term T10.
• The first term a and common difference d are real numbers.

Concept / Approach:
In an arithmetic progression, the nth term is given by Tn = a + (n − 1) * d. We can write equations for T4 and T15 using this formula. The two equations in the two unknowns a and d can then be solved simultaneously. Once a and d are known, T10 follows immediately from the same general term formula. The process is systematic and based on linear equations.

Step-by-Step Solution:
Step 1: Let a be the first term and d be the common difference. Step 2: The fourth term is T4 = a + 3d. We are given T4 = 15, so a + 3d = 15. Step 3: The fifteenth term is T15 = a + 14d. We are given T15 = −29, so a + 14d = −29. Step 4: Now subtract the first equation from the second to eliminate a: (a + 14d) − (a + 3d) = −29 − 15. Step 5: Simplify both sides: a cancels out, and 14d − 3d = 11d on the left, while −29 − 15 = −44 on the right. Step 6: So 11d = −44, leading to d = −44 / 11 = −4. Step 7: Substitute d = −4 into the equation a + 3d = 15. That gives a + 3 * (−4) = 15, so a − 12 = 15. Step 8: Solve for a: a = 15 + 12 = 27. Step 9: Now find the tenth term T10 using T10 = a + 9d. Step 10: Substitute a = 27 and d = −4 to get T10 = 27 + 9 * (−4) = 27 − 36 = −9. Step 11: Therefore, the tenth term of the progression is −9.

Verification / Alternative check:
To verify, check both original conditions with a = 27 and d = −4. The fourth term is 27 + 3 * (−4) = 27 − 12 = 15, which matches the given T4. The fifteenth term is 27 + 14 * (−4) = 27 − 56 = −29, matching the given T15. Since these values of a and d satisfy both conditions, the computation of T10 as −9 is reliable. You can also list a few terms around the tenth term to confirm the pattern decreases by 4 each time and passes through the correct values.

Why Other Options Are Wrong:
Options like −5, −13, −17, or −1 correspond to different assumed values of a and d that do not satisfy both T4 = 15 and T15 = −29. For example, if T10 were −5, then the sequence would not decrease by a consistent common difference that fits the given fourth and fifteenth terms. Only −9 arises naturally from solving the equations formed from the correct arithmetic progression formula.

Common Pitfalls:
A common mistake is to misuse the formula and write Tn = a + nd instead of a + (n − 1) * d, which shifts all term indices and leads to incorrect equations. Another pitfall is arithmetic errors when subtracting the equations or when multiplying d by the required factor to compute T10. Careful step by step algebra and a quick check of the original terms helps avoid these mistakes. Always verify that your found a and d reproduce all given term values before trusting the computed term.

Final Answer:
The value of the tenth term of the arithmetic progression is -9.