Brent@GMATPrepNow wrote:Here's a question I just made up. I'd place it around the 650-700 difficulty level
In a certain sequence, the term an is given by the formula an = (2k + n)/2, where k is a constant. If the sum of all the terms from a1 to a20 inclusive equals 35, then what is the value of k?
A) -3.5
B) - 3.2
C) -3
D) -2.8
E) -2.5
Answer:
A
For any evenly spaced set:
Sum = (number of terms)(average of first and last).
We can PLUG IN THE ANSWERS, which represent the value of k.
Since B and D are messy values, start with C.
C: k = -3, implying that an = (-6 + n)/2
a� = (-6+1)/2 = -5/2 = -2.5.
aâ‚‚ = (-6+2)/2 = -4/2 = -2.
a₃ = (-6+3)/2 = -3/2 = -1.5.
.
.
.
aâ‚‚â‚€ = (-6+20)/2 = 14/2 = 7.
The result is the following evenly spaced set:
(-2.5) + (-2) + (-1.5) + ... + 7.
Sum = (number of terms) (average of first and last) = 20 * (-2.5 + 7)/2 = (10)(4.5) = 45.
Since the sum is too great, the value of k must be SMALLER.
Eliminate C, D and E.
A: k = -3.5, implying that an = (-7 + n)/2
a� = (-7+1)/2 = -6/2 = -3.
aâ‚‚ = (-7+2)/2 = -5/2 = -2.5.
a₃ = (-7+3)/2 = -4/2 = -2.
.
.
.
aâ‚‚â‚€ = (-7+20)/2 = 13/2 = 6.5.
The result is the following evenly spaced set:
(-3) + (-2.5) + (-2) + ... + 6.5.
Sum = (number of terms) (average of first and last) = 20 * (-3 + 6.5)/2 = (10)(3.5) = 35.
Success!
The correct answer is
A.
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