In an arithmetic progression, the 4th term is 11 and the 7th term is -4. What is the value of the 15th term of this arithmetic progression?

Introduction / Context:
This question examines understanding of arithmetic progressions and how to derive unknown terms when two non-consecutive terms are given. It is a standard application of the nth term formula and solving simultaneous linear equations, which makes it a typical aptitude problem on sequences and series.

Given Data / Assumptions:

• The sequence is an arithmetic progression.
• The 4th term is 11.
• The 7th term is -4.
• We must find the 15th term.
• Let a denote the first term and d denote the common difference of the arithmetic progression.

Concept / Approach:
For an arithmetic progression, the nth term is Tn = a + (n - 1) * d. Given T4 and T7, we can form two equations in a and d. Subtracting these equations lets us find d, and then substituting back gives us a. After we know a and d, we use the same formula to calculate T15. This illustrates how relationships between different terms in an arithmetic progression can be exploited to find any term in the sequence.

Step-by-Step Solution:
Step 1: Let a be the first term and d be the common difference.Step 2: The 4th term is T4 = a + (4 - 1) * d = a + 3d. We are given T4 = 11, so a + 3d = 11.Step 3: The 7th term is T7 = a + (7 - 1) * d = a + 6d. We are given T7 = -4, so a + 6d = -4.Step 4: Subtract the first equation from the second to eliminate a: (a + 6d) - (a + 3d) = -4 - 11.Step 5: This simplifies to 3d = -15, so d = -5.Step 6: Substitute d = -5 into a + 3d = 11. Then a + 3 * (-5) = 11, so a - 15 = 11 and a = 26.Step 7: The 15th term is T15 = a + (15 - 1) * d = a + 14d.Step 8: Substitute a = 26 and d = -5: T15 = 26 + 14 * (-5) = 26 - 70 = -44.

Verification / Alternative check:
We can verify by listing some terms. Starting from a = 26 and d = -5, the terms are: T1 = 26, T2 = 21, T3 = 16, T4 = 11, T5 = 6, T6 = 1, T7 = -4. These match the given 4th and 7th terms. Continuing, T8 = -9, T9 = -14, T10 = -19, T11 = -24, T12 = -29, T13 = -34, T14 = -39, T15 = -44, which confirms the answer.

Why Other Options Are Wrong:
Option -49: This would correspond to a different common difference and would not keep T4 and T7 equal to 11 and -4.
Option -39: This is actually T14 in this progression, not T15.
Option -34: This is T13, so it is not the requested term either.
Option -29: This equals T12, again not matching the index n = 15.

Common Pitfalls:
Students sometimes write Tn = a + nd instead of a + (n - 1) * d, which shifts all indices and leads to incorrect results. Others may incorrectly subtract equations, introducing sign errors. It is also easy to miscount when moving from one term to another without relying on the general term formula. Using the formula systematically avoids most counting mistakes.

Final Answer:
The 15th term of the arithmetic progression is -44.