If the sum of the consecutive integers from –42 to n

hi,
how do we determine below? or how do we know that we need to stop at 50 to get 372

Sum up 43 to 50=372
numbers can be weird and it may go upto 43 to 93 or any other number...

Give it a little more thought. How can adding consecutive numbers from 43 to 93 give you a total of 372? In this context, 372 is not a big numbers. You just need a few numbers from 43 onwards to make a sum of 372.

43 + 44 + 45 + 46 + 47 + 48 ... adding a few numbers with an approximate average of 45 gives us a sum of 372.
45*8 = 360 which is close to 372.

So 43 onwards we need only 8 numbers: 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50

# 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.



thanks Bunue,,, I was looking for the # of term formula, because it is not in the Math Book.

Bunuel will you be kind enough to explain this manipulation, I can't seem to get it done!
\((n+43)*(n-42)=744\)

Thanking you in advance.

Multiply \(372=(n+43)*\frac{(n-42)}{2}\) by 2 to get \(744=(n+43)*(n-42)\).

Using AP formula

Sn =\(\frac{n}{2}(2a+(n-1)d)\)
\(372\) = \(\frac{n}{2}(2(-42) + n-1)1)\)
\(n^2-85n-744\)\(=0\)
\((n-93)(n+8)\) \(=0\)
\(n =93 or n = -8\)

What am i missing?

Everything till here is correct. n cannot be negative so n must be 93. So basically we are adding 93 terms starting from -42 to get the sum of 372.


-42 -41 - 40 .... 93 terms
First term - 42
Second term - 41 (you obtain by adding 1 to -42)
Third term -40 (you obtain by adding 2 to -42)
93rd term 50 (you obtain by adding 92 to -42)
So last term is 50.

OR

Last term = a + (n-1)*d = -42 + (93 - 1)*1 = 50

please clarify the method the sum of consecutive integer when the first term is not 1

i did not find this rule in you math book, please clarify the rule

Post your query once. You posted 2 related posts in less than 30 minutes.

As for your question, sum of n terms of an arithmetic progression (or a sequence in which the difference between consecutive terms is equal) is given by:

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

where n= number of terms, a=first term, d=difference between consecutive terms

Alternately, the same rule becomes , S= (first term+last term)/2 or in other words, the sum is equal to the average of the first and the last term of an arithmetic progression.

My Solution:

Digits from -42 to + 42 will be cancelled out, therefore there sum becomes 0 and first digit would me 43.

Minimum value of "n" as per answer choices is 47 and maximum value of "n" is 51.

So let's try digit 51,

43+44+45+46+47+48+49+50+51=423, but required sum is 372.

Let's try 50, 43+44+45+46+47+48+49+50=372, Answer is Option D

For addition we can also use formula [b](Last term + first term)/2 * Number of terms[/b]

My Solution:

Digits from -42 to + 42 will be cancelled out, therefore their sum becomes 0 and first digit would me 43.

Minimum value of "n" as per answer choices is 47 and maximum value of "n" is 51.

So let's try digit 51,

43+44+45+46+47+48+49+50+51=423, but required sum is 372.

Let's try 50, 43+44+45+46+47+48+49+50=372, Answer is Option D

For addition we can also use formula [b](First term + Last term)/2 * Number of terms[/b][/b]

2 methods...
we know for sure that n must be > 42.
we can test numbers.
so -42 up to 42, will equal to 0.
now we have:
43, 44, 45, 46, 47, 48, 49, 50, 51 - let's test...
43+47 = 90
44+46 = 90
45+48 = 93
total: 273..we are short 99..which must be 49 and 50..so n must be 50.

second approach...
find sum of the first 42 numbers
we can do so by applying the formula: n(n+1)/2.
42*43/2 = 21*43 = 903. but since we have -42, it must be true that this is -903.
now..n is greater than 42...and when the sum of all positive and negative is 372, then the sum of only positive numbers would be: 903+372 = 1275.
n(n+1)=1275
since we need a number that either ends in 5 or if divided by 2 has the last digit 5, we an eliminate
A (has 7 as units digit, and 48 - 8 as units digit)
B (has 8 as last digit, and 49 - 9 as last digit)
C (has 9 as last digit, and 50/2 - 5 as last digit) - hold
D if divided by 2 has 5 as the last digit - hold
E no.

now..between C and D...it doesn't take long to see that C is not working...
49x50/2 = 49x25 = (50-1)x25 = 1250 -25 = 1225. but we need 1275. so out.

D - 50x51/2 = 25*51 = 1275.

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

Another approach, process elimination:

Let's try D 50

Sum = 50 - -42 = 92 + 1 = 93

S = n/2(a+l)
93/2(50 - 42)
41.5(8)
= 372

Therefore, the answer is D

My approach:
-42 + -41 + .... + 42 = 0
43 + 44 = 87 + 45 = 132 + 46 = 178 + 47 = 225 + 48 = 273
273 + 49 = 322 + 50 = 372

Therefore, the answer is D

Pretty interesting question
Here's how i solved it
sum of -42 to 42 is equal to 0
Now start from 43
Start plugging in the answer choices
Start with c

43 to 49
Sum from 43 to 49
Number of integers from 43 to 49
=49-43=6+1=7
Sum=7*(49+43/2)
=7*46=322

Since 322 is lesser than 372 hence options a,b and c can be ruled out.
Since we know that sum of integers from 43 to 49 is 322
If we add 50 to it , we get our answer
322+50=372

Therefore the correct option is "D"

Neglect numbers from -42 to +42, because the summation of this series is positive

Start from 43

\(Sum = [2*43 + (n-1)] * \frac{n}{2}\)
\(744 = 86n + n^2 - n = n^2 +85n\)

\(744 = n^2 + 85n\)

Now substitute : {answer choice - 43 + 1} in above equation

50 it is

OR

\(Sum = (first term + last term) * \frac{no of terms}{2}\)

RHS = (43 + answer choice) * (answer choice - 43 + 1)/2