In an arithmetic progression, the 7th term is -15 and the 12th term is 5. What is the value of the 16th term of this arithmetic progression?

Introduction / Context:
This question checks understanding of arithmetic progressions, especially how to use information about two non-consecutive terms to find any other term. In aptitude exams and competitive tests, candidates are often required to manipulate the general term formula of an arithmetic progression to obtain unknown values such as the first term, common difference, or any specific nth term.

Given Data / Assumptions:

• The sequence is an arithmetic progression (A.P.).
• The 7th term of the A.P. is -15.
• The 12th term of the A.P. is 5.
• We are asked to find the 16th term of the same A.P.
• Let a be the first term and d be the common difference.

Concept / Approach:
For an arithmetic progression, the nth term is given by the formula Tn = a + (n - 1) * d. When we know two terms, say Tm and Tn, we can form two equations in a and d. Solving these equations simultaneously gives the values of a and d. Once a and d are known, we can compute any required term such as T16 using the same nth term formula. The problem is a straightforward example of forming and solving linear equations.

Step-by-Step Solution:
Step 1: Let a be the first term and d be the common difference.Step 2: The 7th term is T7 = a + (7 - 1) * d = a + 6d. We are given T7 = -15, so a + 6d = -15.Step 3: The 12th term is T12 = a + (12 - 1) * d = a + 11d. We are given T12 = 5, so a + 11d = 5.Step 4: Subtract the first equation from the second: (a + 11d) - (a + 6d) = 5 - (-15).Step 5: This simplifies to 5d = 20, so d = 4.Step 6: Substitute d = 4 into a + 6d = -15 to find a: a + 6 * 4 = -15, so a + 24 = -15 and a = -39.Step 7: The 16th term is T16 = a + (16 - 1) * d = a + 15d.Step 8: Substitute a = -39 and d = 4: T16 = -39 + 15 * 4 = -39 + 60 = 21.

Verification / Alternative check:
We can list a few terms around the 7th and 12th positions to verify. Starting with a = -39 and d = 4, the terms are: T1 = -39, T2 = -35, T3 = -31, T4 = -27, T5 = -23, T6 = -19, T7 = -15, T8 = -11, T9 = -7, T10 = -3, T11 = 1, T12 = 5. These match the given values. Continuing forward, T13 = 9, T14 = 13, T15 = 17, T16 = 21, which confirms our answer.

Why Other Options Are Wrong:
Option 25: This would correspond to a different common difference, inconsistent with T7 and T12.
Option 29: This value would require the common difference to be higher than 4, which would break the given term values.
Option 33: This is larger still and would make T7 and T12 incorrect if we recalculated them.
Option 17: This is smaller than the correct value and would not satisfy the pair of equations derived from the 7th and 12th terms.

Common Pitfalls:
Candidates sometimes use the wrong index in the nth term formula, writing Tn = a + nd instead of a + (n - 1) * d. Another frequent mistake is to directly subtract the term values without forming proper equations, or to mix up which equation to subtract from which, leading to sign errors. Careful algebra and clear identification of T7 and T12 are important to avoid mistakes.

Final Answer:
The 16th term of the arithmetic progression is 21.