In an arithmetic progression, the 3rd term is 13 and the 6th term is -5. What is the 11th term of this progression?

Introduction / Context:
This question involves an arithmetic progression (A.P.), which is a sequence of numbers where the difference between consecutive terms is constant. You are given the 3rd and 6th terms of an A.P. and are asked to find the 11th term. The problem assesses your ability to apply the general n-th term formula of an A.P. and solve for both the first term and the common difference using the information provided.

Given Data / Assumptions:

- Let the first term be a. - Let the common difference be d. - The 3rd term T_3 is 13. - The 6th term T_6 is -5. - We need to determine the 11th term T_11.

Concept / Approach:
The n-th term of an arithmetic progression is given by T_n = a + (n - 1)d. Using this formula, T_3 can be written as a + 2d, and T_6 as a + 5d. Since their values are known, we get two linear equations in a and d. Solving these equations gives us specific values of a and d. Then we substitute these into the formula for T_11, which is a + 10d. This straightforward use of the general term formula is the standard way to approach such problems.

Step-by-Step Solution:
Step 1: Write T_3 in terms of a and d: T_3 = a + (3 - 1)d = a + 2d. Step 2: We are told T_3 = 13, so a + 2d = 13. Call this Equation (1). Step 3: Write T_6 in terms of a and d: T_6 = a + (6 - 1)d = a + 5d. Step 4: We are told T_6 = -5, so a + 5d = -5. Call this Equation (2). Step 5: Subtract Equation (1) from Equation (2): (a + 5d) - (a + 2d) = -5 - 13. Step 6: This simplifies to 3d = -18, so d = -6. Step 7: Substitute d = -6 back into Equation (1): a + 2 * (-6) = 13. Step 8: So a - 12 = 13, hence a = 25. Step 9: Now find T_11 using T_11 = a + (11 - 1)d = a + 10d. Step 10: Substitute a = 25 and d = -6: T_11 = 25 + 10 * (-6) = 25 - 60. Step 11: Compute 25 - 60 = -35, so T_11 = -35.

Verification / Alternative check:
To verify, write the first few terms using a = 25 and d = -6: 25, 19, 13, 7, 1, -5, -11, and so on. The 3rd term is 13 and the 6th term is -5, as required. From the 1st term to the 11th term, we move 10 steps, each subtracting 6, so T_11 = 25 - 10 * 6 = -35. This confirms the earlier calculation and shows the progression is consistent with the given data.

Why Other Options Are Wrong:
Values such as -29, -41 and -47 result from incorrect differences or miscounting the number of steps between terms. For example, using the wrong formula a + nd or misapplying the subtraction of equations can produce such incorrect results. Only -35 satisfies the progression defined by T_3 = 13 and T_6 = -5 with a constant difference of -6.

Common Pitfalls:
Students sometimes mistakenly write the general term as a + nd instead of a + (n - 1)d, which shifts all term indices and leads to incorrect answers. Others may mis-handle the signs when subtracting the equations for T_3 and T_6. Carefully writing each equation with full detail and double checking the subtraction step helps avoid these common issues.

Final Answer:
The 11th term of the arithmetic progression is -35, which corresponds to option D.