In an arithmetic progression, the sum of the 3rd term and the 15th term is equal to the sum of the 6th, 11th and 13th terms. Which term of the progression must be equal to zero?

Introduction / Context:
This question involves properties of arithmetic progressions and how specific terms can be related using algebra. You are asked to use the definition of general terms of an arithmetic progression and a condition on the sum of several terms, and then deduce which term must be zero. This is a classic reasoning pattern for arithmetic progression questions in aptitude and entrance exams.

Given Data / Assumptions:

• We have an arithmetic progression with first term a and common difference d.
• Sum of the 3rd and 15th terms equals the sum of the 6th, 11th and 13th terms.
• The terms are indexed in the standard way: Tn = a + (n - 1) * d.
• We must find which numbered term of the progression is necessarily equal to zero.

Concept / Approach:
We express each mentioned term in terms of a and d. Then we use the given equality of sums to obtain an equation relating a and d. From that equation we find a linear relation between a and d. Finally, we determine for which term number k the expression a + (k - 1) * d becomes zero.

Step-by-Step Solution:
Step 1: Write required terms: T3 = a + 2d, T15 = a + 14d.Step 2: Their sum is T3 + T15 = (a + 2d) + (a + 14d) = 2a + 16d.Step 3: Write other terms: T6 = a + 5d, T11 = a + 10d, T13 = a + 12d.Step 4: Their sum is T6 + T11 + T13 = (a + 5d) + (a + 10d) + (a + 12d) = 3a + 27d.Step 5: Given that the two sums are equal, so 2a + 16d = 3a + 27d.Step 6: Rearrange: 0 = a + 11d, so a = -11d.Step 7: General term Tn = a + (n - 1)d. Substitute a = -11d: Tn = -11d + (n - 1)d = (n - 12) * d.Step 8: For Tn to be zero with d not equal to zero, we require n - 12 = 0, thus n = 12.

Verification / Alternative check:
If T12 is zero, then Tn = (n - 12) * d. You can quickly check that T3 + T15 equals T6 + T11 + T13 when you express each as multiples of d. This confirms that the 12th term must be zero whenever the condition holds, independent of the actual value of d (except d not equal to zero).

Why Other Options Are Wrong:
The 7th, 9th, 14th or 15th terms would require different values of n and would make Tn zero at a different index, which would break the derived relation a = -11d. With any of those indices, the equality between the two sums would not hold for a non zero progression.

Common Pitfalls:

• Writing incorrect indices for the terms, for example mixing T3 with a + 3d instead of a + 2d.
• Forgetting that the common difference d cannot be zero if the problem is to be meaningful.
• Trying to plug in arbitrary numbers instead of using the general term form and algebra.

Final Answer:
The term that must be equal to zero is the 12th term of the arithmetic progression.