How to manipulate standard formulas

We know the formula we need to use to find the sum of n consecutive positive integers starting from 1. The formula is given as n(n+1)/2. So the sum of first four positive integers is 4 * (4 + 1)/2 = 10. This might seem a bit cumbersome, since it is easy to see that 1 + 2 + 3 + 4 = 10, but we know that the formula comes in very handy when n is a large number. For example, the sum of first 50 positive integers = 50 * 51/2 = 1275. Obviously, this will be a lot harder when done the “1 + 2 + 3 + 4 … + 49 + 50” way. Now the question is, how do we adjust the same formula to find the sum of consecutive integers which do not start from 1? Say, how do we find the sum of all positive integers from 8 to 20? The formula assumes a starting point of 1, so then we insert only the last number, n. How do we manage the 8? Let’s try to figure it out Say the sum of first 20 positive integers = 1 + 2 + 3 + 4 + …. + 19 + 20 = 20 * 21/2 (1 + 2 + 3 +… + 7) + (8 + 9 +10 + … + 19 + 20) = 20 * 21/2 We need the value of the part in red, let’s call it the required sum. (1 + 2 + 3 +… + 7) + The Required Sum = 20 * 21/2 Note here that we know the sum of 1 + 2 + 3 + … + 7 = 7 * 8/2 So, 7*8/2 + The Required Sum = 20 * 21/2, therefore the Required Sum = 20*21/2 – 7*8/2 To get the sum of consecutive integers from 8 to 20, we find the sum of all integers from 1 to 20 (using the formula we know) and subtract the sum of integers from 1 to 7 out of it (using the same formula). To generalize, the sum of all positive integers from m to n is given as: n(n+1)/2 – (m-1)*m/2 Let’s look at a question based on this concept: If the sum of the consecutive integers from –40 to n inclusive is 356, what is the value of n? (A) 47 (B) 48 (C) 49 (D) 50 (E) 51 If you are thinking that we haven’t gone over how to adjust the formula for negative numbers, you are right, but what we have discussed is enough to solve this question. Numbers around 0 are symmetrical. So 1 and -1 add up to equal 0. Similarly, 2 and -2 add up to equal 0, and so on… -40, -39 … 0 … 39, 40, 41, 42, 43, 44, 45 … The sum of all numbers from -40 to 40 will be 0. Or another way to look at it is that 0 is the mean of all numbers from -40 to 40. So the total sum of these numbers will also be 0. The given sum is actually the sum of numbers from 41 to n only. We know how to calculate that: n(n+1)/2 – 40*41/2 = 356 n(n+1) = 2352 From the options, we see that n cannot be 49 or 50 because the product of 49*50 or 50*51 will end in 0, so plug in n = 48 to check whether 48*49 is equal to 2352. It is, therefore our answer is B (Had we obtained a lower product than required, we could have said that n must be 51. Had we obtained a higher product than was required, we could have said that n is 47.) Another method: Use the concept of arithmetic mean and ballpark. The mean of numbers from 41 to 47 or 48 or 49… will be somewhere between 44 and 46. Let’s estimate the number of integers we need to get the sum of about 356. Each additional integer is quite large (more than 45) therefore, a difference of about 10-15 in the sum due to the various possible values of the mean will be immaterial. 45*7 = 315 45*8 = 360 This brings us very close to the value of 356. Assuming there are 8 integers, their values will be from 41 to 48. The average of these 8 numbers will be 44.5. The total sum will be 44.5 * 8 = 356. It matches, so our answer is still B.