The sum of the consecutive integers from -34 to n inclusive is 108. what is the value of n?

Question

Im having such a hard time with this

Answer 1

Start by noting that n will certainly be a positive integer. The sum of integers from -34 to 0 will of course be negative, so you need to add a lot of positive integers to get a large positive number like 108.

You need to make some use of the formula Sn = n(n+1)/2, where Sn is the sum of consecutive integers from 1 to n. The sum from -34 up to -1 will be equal to -S34 = -(34)(35)/2 = -595. That means you need to add positive integers from 1 to n that actually has a sum of 595 + 108 = 703. So you need to find n such that n(n+1)/2 = 703 ==> n(n+1) = 1406 ==> n^2 + n - 1406 = 0. The quadratic formula tells us that n = (-1 + sqrt(1 + 4*1406))/2 = (-1 + sqrt(5625))/2 = (-1 + 75)/2 = 74/2 = 37. So n = 37.

There is another, less analytical way to solve this. Starting from -34, you can see that if you add every number from -34 up to -1 and then from 1 up to 34, you'll be back at zero. Then you can just sum the subsequent numbers to see what you get.

0 + 35 = 35
35 + 36 = 71
71 + 37 = 108

This confirms the result of n = 37.

Answer 2

The consecutive integers from -34 to n represent an arithmetic sequence.

All you have to do is get the formula for the arithmetic series where the first term is equal to -34 and the difference is 1.

The formula for an arithmetic series is

S = n[ 2a + (n - 1)d ] / 2

In our case, a = -34, and d = 1. Equating this to 108, we get

108 = n[ 2(-34) + (n - 1)1 ] / 2
108 = n[ -68 + n - 1] / 2
108 = n[ -69 + n] / 2
216 = n[ -69 + n]
216 = -69n + n^2
0 = n^2 - 69n - 216

Factoring this,

0 = (n - 72)(n + 3)

Therefore, n = {72, -3} (showing we have two values for n).

Let's check:

Sum of integers from -34 to 72:
a = -34, n = 72, d = 1:

S = n[ 2a + (n - 1)d ] / 2
S = 72[2(-34) + (72 - 1)(1)] / 2
S = 72[-68 + 71] / 2
S = 72[3]/2
S = 216/2 = 108, so this checks out.

For some reason, n = -3 is extraneous and will not work.

Answer 3

ok, reason as follows:

you probably already know that the sum from 1 to n is n(n+1)/2. So, that's where we start.

Count all the numbers from 1 to n, but we've include too many so we'll subtract them off:

1 + 2 + 3 + ... + 34 + 35 + 36 + ... + n = n(n+1)/2

but we don't want 1 + 2 + 3 +...+ 34 = 34*35/2 so we'll subtract it off:

n(n+1)/2 - 34*35/2 = 108

n^2 + n - 1406 = 0

you'll find:

n = -38
and
n= 37

can't be -38 so:

n = 37

check 35 + 36 + 37 =108

Answer 4

The answer is 37 because numbers -34 through 34 = 0. 35+36+37=108

Answer 5

You know that the sum of number -34 to 0 will cancel the sum from 0 to 34 so just add the successive integers following 34.
35 + 36 = 71 + 37 = 108, n = 37!

Answer 6

-34 to +34 adds to 0
then 35,36,37 adds to 108.
so n= +37

Answer 7

If a+j=p_1^a_1*p_2^a_2*...*p_r^a_r is the top factorization of a+j, with each and every a_i>0, then a+j might have precisely (a_1+a million)*(a_2+a million)*...*(a_r+a million) divisors, which comprise a million and a+j. subsequently interior the present subject, we would desire to have (a_1+a million)*(a_2+a million)*...*(a_r+a million) =6. this implies r<=2. In case r=a million, we would desire to have a_1=5 and subsequently a+j=p^5 for some top p. In case r=2, (a_1,a_2)=(a million, 2) or (2,a million), so for this reason we would desire to have a+j=p*q^2 for some primes p and q. subsequently the sequence a+a million, a+2, .., a+N might desire to comprise basically integers which could be factored into between the types p^5 or p*q^2.

Answer 8

-3 and 72

Answer 9

maybe because the problem is incomplete.
how many consecutive intergers are you talking? 2, 3 , 4 or what?

Answer 10

yes