Find m if the sum of the consecutive integers from -16 to m, inclusive

Done. Thank you for the suggestion.

Imo D
Can be calculated from sum formula of a sequence

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Opt D
Other way round
S = n/2(2a+(n-1)d)
=> 74=n/2(-32+n-1) as common difference is 1
148=-33n+n^2
n^2-33n-148=0
upon simplification n = 37
tn=a+(n-1)d
m=16+(37-1)1
m=20

We can use the following formula:

sum = average x quantity

Since we have a set of evenly spaced integers, the average is (-16 + m)/2.

We can also calculate the quantity as m - (-16) + 1 = m + 17.

Thus:

74 = (-16 + m)/2 x (m + 17)

148 = (m - 16)(m + 17)

148 = m^2 + m - 272

m^2 + m - 420 = 0

(m + 21)(m - 20) = 0

m = -21 or m = 20

Since m > -16, m must be 20.

Alternate Solution:

The integers in the sum will be -16, -15, -14, …, -2, -1, 0, +1, +2, …, m. Notice that all the negative integers from -16 to -1 inclusive will be offset by the positive integers from +1 to +16, yielding a net sum of 0. So now, starting with the integer +17, we need to add consecutive integers until we get a sum of 74. We see that 17 + 18 + 19 + 20 = 74, and thus m = 20, the largest integer in the set.

Answer: D

[(-16+m)/2](17+m)=74→
m^2+m-420=0→
x=20
D

That's take me too long about 10 mins to prove it's D. I need more practice.

First term: -16 and Last term : m

Sum: 74

Sum from -16 to +16 will be '0', hence to get 74 as the sum, we will have to extend the series ahead.

17 + 18 + 19 + 20 = 74.

Therefore, m = 20

Answer D

All terms till 16 will cancel out.

Now we have 17 + 18 +19 ...n terms = 74
=17 +17 +1 +17+2....n terms
=17n +n(n-1)/2 = 74 (note we have n-1 terms 1,2,3...)
34n +n^2 -n = 148
(n+34)(n-4) =0
n = 4
there are 4 terms (including 17)
17+3 = 20, Answer