Difficulty: Medium
Correct Answer: 1962
Explanation:
Introduction / Context:
This problem is a standard arithmetic progression (A P) question where you need to find the sum of a specified number of terms. Mastering such questions is essential for aptitude exams because they combine formula recall with careful substitution and arithmetic.
Given Data / Assumptions:
Concept / Approach:
For an arithmetic progression, the nth term is given by a_n = a + (n − 1) * d. The sum of the first n terms is S_n = n * (a + a_n) / 2. So we first find the 36th term using the nth term formula and then apply the sum formula.
Step-by-Step Solution:
Step 1: Compute the 36th term a_36.
a_36 = a + (n − 1) * d = 2 + (36 − 1) * 3 = 2 + 35 * 3.
Step 2: Evaluate 35 * 3 = 105, so a_36 = 2 + 105 = 107.
Step 3: Use the sum formula S_36 = n * (a + a_36) / 2.
S_36 = 36 * (2 + 107) / 2 = 36 * 109 / 2.
Step 4: Simplify: 36 / 2 = 18, so S_36 = 18 * 109.
Step 5: Multiply 18 * 109 = 18 * 100 + 18 * 9 = 1800 + 162 = 1962.
Verification / Alternative check:
As a quick check, observe that the average of the first and last terms is (2 + 107) / 2 = 54.5. In any A P the average term equals this midpoint value. The sum of 36 terms is then 36 * 54.5, which again gives 1962. This cross check confirms that the calculation is consistent.
Why Other Options Are Wrong:
The other options 3924, 1684, and 1452 come from typical mistakes such as forgetting to divide by 2 in the sum formula, miscomputing the 36th term, or using an incorrect value for n. None of them match the correctly derived sum of 1962.
Common Pitfalls:
Candidates sometimes mix up the formulas for the nth term and the sum, or they use n instead of (n − 1) when computing a_n. Another common error is arithmetic slipping when multiplying larger numbers. Writing each step clearly and simplifying fractions early reduces the chance of mistakes.
Final Answer:
The sum of the first 36 terms of the progression 2, 5, 8, 11, ... is 1962.
Discussion & Comments