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

this problem should be solved normally by using the following formula: k(k+1)/2

Using the equation below: n*(n+1)/2 - 42*43/2 = 372 It is much easier to get n*(n+1) = 2550 => n=50

sum of integers from -42 to 0 inclusive=(-42+0)/2*43=-903 372-(-903)=1275=sum of integers from 1 to n inclusive 1275=(n+1)/2*n→ n^2+n-2550=0→ (n+51)(n-50)=0 n=50 D

target is to determine sum from -42 to n ; -42 to 42 will be zero so sum 0 to 42 ; 42*42 /2 ; 903 and use answer options sum to 50 ; 50*51/2 ; 1275 1275-903 ; 372 ; which matches our target sum from -42 to n OPTION D

Another approach is to recognize that -42 to n results in a positive number. So all the negatives will cancel out up to 42, in which case you will only add the positive numbers. 43+44+45+46+47+48+49+50 = 372

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 -> correct : till +42, the sum 0, so from 43 to n the sum =372 i.e. (43+n)*(n-43+1)/2=372 => (n+43)*(n-42)=372*2 => n =50 (plugging the values from answer choices) E. 51

This problem can be solved in an easier way I couldn't find in the answer choices. Sometimes it difficult to put in formulas and find the answers. How I approached the problem can be done under 2 mins. It took me 1 min 30 secs: -42 to +42 = 0 Add all the numbers after that. Not a lot to add. 43+44+45+46+47+48+49+50 = 372 Saves time. If the number is bigger than this then we can apply the mean formulas.

always avoid long calculation.. when we see 744 break it down to simple factors.. 2x4x3x31... now since (n+43)(n-42) = 2x4x3x31 .. 23 can see 2x4 = (50-42)... test 50 on (n+43) = u see 93 ( 3x31)

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