jjack0310 wrote:
Bunuel wrote:
gmattokyo wrote:
If the sum of the consecutive integers from 42 to n inclusive is 372, what is the value of n?
A. 47
B. 48
C. 49
D. 50
E. 51
A. 47
B. 48
C. 49
D. 50
E. 51
# of terms =42+1+n=(n+43)
Sum=372=(n+43)*\frac{(n-42)}{2}
744=(n+43)*(n-42)
n=50
OR
42 terms after zero and 42 terms below zero will total 0. So, our new question will be consecutive integers with first term 43 have sum 372, what is the last term:
\frac{43+n}{2}*(n-43+1)=372
(n+43)*(n-42)=744
n=50
Answer: D (50)
Can you please explain how is the number of terms 42 + n +1?
Thanks
# of term from a to b, inclusive is b-a+1.
# of term from -42 to n, inclusive is n-(-42)+1=42+n+1.