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Re: Sum of consecutive integers
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23 Jul 2013, 14:42



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Re: If the sum of the consecutive integers from –42 to n
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30 Sep 2013, 21:45
It is consecutive integers, So, 42...41...40....1...0...+1...+2....+42...(+43...+44....+49...+50) Sum up 43 to 50=372 n=50
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Re: Sum of consecutive integers
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01 Oct 2013, 04:04
hi all, it becomes quite an easy question if we know the below basic formulae nth term in A.P = l = a+ (n1)d where l = last number in series, a= first number in series, n = total number of terms in series and d = diff sum of n terms = (a+l)*n/2 Now applying the above a= 42 d =1 Sum = 372 last term = l = 42 + (n1) = n43 sum = (42 +n43) *n/2 = 372 = (n85)*n = 744 =\(n^2\)85n744 = 0 (n93)(n+8) =0 n can't be 8 and hence n =93 hence the last term here = l = a + (n1)d = 42 + (931)*1 = 42+92 = 50
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Re: If the sum of the consecutive integers from –42 to n
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01 Oct 2013, 04:09
monirjewel wrote: It is consecutive integers,
So, 42...41...40....1...0...+1...+2....+42...(+43...+44....+49...+50) Sum up 43 to 50=372 n=50 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=372numbers can be weird and it may go upto 43 to 93 or any other number...
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Re: If the sum of the consecutive integers from –42 to n
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01 Oct 2013, 19:01
ankur1901 wrote: monirjewel wrote: It is consecutive integers,
So, 42...41...40....1...0...+1...+2....+42...(+43...+44....+49...+50) Sum up 43 to 50=372 n=50 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=372numbers 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
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Re: Sum of consecutive integers
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02 Oct 2013, 09:02
# of term from a to b, inclusive is ba+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.
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Re: Sum of consecutive integers
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05 Oct 2013, 17:12
Bunuel wrote: # of terms =\(42+1+n=(n+43)\)
\(Sum=372=(n+43)*\frac{(n42)}{2}\)
\(744=(n+43)*(n42)\)
\(n=50\)
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:
\(\frac{43+n}{2}*(n43+1)=372\)
\((n+43)*(n42)=744\)
\(n=50\)
Answer: D (50)
Bunuel will you be kind enough to explain this manipulation, I can't seem to get it done! \((n+43)*(n42)=744\)Thanking you in advance.



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Re: Sum of consecutive integers
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06 Oct 2013, 02:55



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Re: If the sum of the consecutive integers from –42 to n
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07 Oct 2013, 16:42
Using AP formula
Sn =\(\frac{n}{2}(2a+(n1)d)\) \(372\) = \(\frac{n}{2}(2(42) + n1)1)\) \(n^285n744\)\(=0\) \((n93)(n+8)\) \(=0\) \(n =93 or n = 8\)
What am i missing?



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Re: If the sum of the consecutive integers from –42 to n
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07 Oct 2013, 19:45
GMAT40 wrote: Using AP formula
Sn =\(\frac{n}{2}(2a+(n1)d)\) \(372\) = \(\frac{n}{2}(2(42) + n1)1)\) \(n^285n744\)\(=0\) \((n93)(n+8)\) \(=0\) \(n =93 or n = 8\)
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 + (n1)*d = 42 + (93  1)*1 = 50
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Re: If the sum of the consecutive integers from –42 to n
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04 Oct 2015, 07:48
Bunuel wrote: gmattokyo wrote: Nice solution. Correct as always. I was actually thinking of having a ballpark and reach the answer from 43 to 50. Add from 43, 44, 45.. to reach 372, need at least 7 (43x7=301) and less than 9 (43x9=387).
Did you use any intermediary formulas for this part? (n+43)*(n42)=744 n=50 Sum of consecutive integers or sum of terms in any evenly spaced set (AP): \(\frac{first term+last term}{2}*number of terms\) In original question: First term: 42 Last term: n Number of terms: 42+n+1=n+43 \(\frac{(n42)}{2}*(n+43)=372\) \((n42)*(n+43)=744\) please clarify the method the sum of consecutive integer when the first term is not 1



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Re: If the sum of the consecutive integers from –42 to n
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04 Oct 2015, 07:56
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 # of terms =\(42+1+n=(n+43)\) \(Sum=372=(n+43)*\frac{(n42)}{2}\) \(744=(n+43)*(n42)\) \(n=50\) 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: \(\frac{43+n}{2}*(n43+1)=372\) \((n+43)*(n42)=744\) \(n=50\) Answer: D (50) i did not find this rule in you math book, please clarify the rule



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Re: If the sum of the consecutive integers from –42 to n
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04 Oct 2015, 08:24
anik19890 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 # of terms =\(42+1+n=(n+43)\) \(Sum=372=(n+43)*\frac{(n42)}{2}\) \(744=(n+43)*(n42)\) \(n=50\) 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: \(\frac{43+n}{2}*(n43+1)=372\) \((n+43)*(n42)=744\) \(n=50\) Answer: D (50) 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+(n1)*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.



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Re: If the sum of the consecutive integers from –42 to n
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20 Oct 2015, 10:11
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 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]
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If the sum of the consecutive integers from –42 to n
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20 Oct 2015, 10:16
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 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]
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Re: If the sum of the consecutive integers from –42 to n
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11 Mar 2016, 18:52
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 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 = (501)x25 = 1250 25 = 1225. but we need 1275. so out. D  50x51/2 = 25*51 = 1275.



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Re: If the sum of the consecutive integers from –42 to n
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17 Mar 2017, 10:31
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



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Re: If the sum of the consecutive integers from –42 to n
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25 Mar 2017, 19:31
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



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If the sum of the consecutive integers from –42 to n
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28 Mar 2017, 11:12
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 =4943=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"



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If the sum of the consecutive integers from –42 to n
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01 Apr 2017, 16:45
Neglect numbers from 42 to +42, because the summation of this series is positive Start from 43 \(Sum = [2*43 + (n1)] * \frac{n}{2}\) \(744 = 86n + n^2  n = n^2 +85n\) \(744 = n^2 + 85n\) 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
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