an = (2k + n)/2, where k is a constant


a1 = (2k+1)/2
a2 = (2k+2)/2
a3 = (2k+3)/2
.
.
. observing the trend
.
a20 = (2k+20)/2

Sum of all terms
a1 +a2+a3.....+a20 = (2k+1)/2 + (2k+2)/2 + (2k+3)/2 ..... + (2k+20)/2
35 = ( 20*2K + 1 +2 + 3+4+5+6......+18+19+20)/2


Using the formula #of numbers * avg = sum of numbers,
1+2+3+4....+18+19+20 = 10.5*20=210

putting this back in main question

35 = ( 20*2K + 1 +2 + 3+4+5+6......+18+19+20)/2
35 = ( 20*2K + 210)/2
35 = 20K + 105
35 - 105 = 20K
20K = -70
K = -70 /20
K = -3.5

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.

I think there's an easier way here, just decomposing the fraction.

An = (2k/2) + n/2 = k + n/2

So if we're adding A1 to A20, we've got

k + 1/2 + k + 2/2 + k + 3/2 + ... + k + 20/2

or

20k + (1 + 2 + ... + 20)/2

or

20k + 105

So we've got 35 = 20k + 105, or k = -3.5

Another way would be working with the terms. We know that

An = (2k + n)/2

2*An = 2k + n

2*An - n = 2k

An - (n/2) = k

so our first two terms give

A1 - 1/2 = k
A2 - 1 = k

or

A1 - 1/2 = A2 - 1

A1 = A2 - 1/2

From here, we can see that the sequence will go up by 0.5 each time. That means

A1 + A2 + ... + A20 =

A1 + A1 + .5 + A1 + 1 + ... + A1 + 9.5 =

20A1 + (.5 + ... 9.5) =

20A1 + 95 = 35

20A1 = -60

A1 = -3

From here, we know A1 - 1/2 = k, so k = -3.5.