Solution: To save time and improve accuracy on DS questions in GMAT, learn and apply the Variable Approach.
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
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Now we will solve this DS question using the Variable Approach.
Let’s apply the 3 steps suggested previously.
Follow the first step of the Variable Approach by modifying and rechecking the original condition and the question.
Sequence S has the difference between a term and its previous term constant. Thus, it is an arithmetic sequence.
The general term of an arithmetic sequence Tn = a + (n-1)d – where ‘a’ is the first term, n is total terms and d is a common difference.
We have n = 250. Thus, \(T_{250}\) = a + (250 - 1)d = a + 249d.
We have to find n= 200 or a + 199d.
Follow the second and the third step: From the original condition, we have 2 variables (‘a’ and d). To match the number of variables with the number of equations, we need 2 equations. Since conditions (1) and (2) will provide 1 equation each, C would most likely be the answer.
Recall 3- Principles and Choose C as the most likely answer. Let’s look at both conditions combined together.
Condition (1) tells us that the 150th term of Sequence S is 305: 305 = a +149d – Equation (1)
Condition (2) tells us that the 100th term of Sequence S is −95: -95 = a + 99d – Equation (2)
Equation (1) – Equation (2)
=> 305 – (- 95) = a + 149d – ( a + 99d)
=> 400 = 50d
=> d = 4
Substituting d = 4 into equation (1):
=> 305 = a + 149(4)
=> 305 = a + 596
=> 305 – 596 = a
=> a = -291
Therefore, \(200^{th}\) term a + 199d = -291 + 199(4) = -291+796 = 505
The answer is a unique value; both conditions combined are sufficient according to Common Mistake Type 2 which states that the answer should be a unique value.
Both conditions together are sufficient.
Therefore, C is the correct answer.
Answer: C
