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If the sum of all the even numbers from 1 to n(n: odd number) is 69*70
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18 Jul 2017, 01:11
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If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n? A. 135 B. 137 C. 139 D. 141 E. 143
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If the sum of all the even numbers from 1 to n(n: odd number) is 69*70
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18 Jul 2017, 17:17
MathRevolution wrote: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n?
A. 135 B. 137 C. 139 D. 141 E. 143 1. If you don't remember the formula for summing consecutive even integers, you can use the general formula for the sum of any evenly spaced set where Sum = average (arithmetic mean) * number of termsThe average (arithmetic mean) of evenly spaced consecutive numbers is First + Last divided by 2. 2. Given 69*70, from the basic formula you know that:  one of those numbers is the average of the first and last terms, and  the other number is the number of termsThe first even number from 1 to n (where n is odd) is 2. Looking again at 69*70 . . . the average of two even terms (first and last) cannot be odd. So 69 must be the number of terms. And 70 must be the average of 2 and the last even term. 3. \(\frac{2 + Last Even}{2}\)= 70 2 + last even term = 140 Last even term = 138. n is odd. For 138 to be included in 1 to n . . . n must be 139. (Or, from the outset it should be clear that the last even term is n  1.) Answer C
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If the sum of all the even numbers from 1 to n(n: odd number) is 69*70
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18 Jul 2017, 01:38
MathRevolution wrote: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n?
A. 135 B. 137 C. 139 D. 141 E. 143 Sum of first \(n\) positive even numbers is given by formula \(= n(n+1)\)
Sum of all even numbers from \(1\) to \(n = 69 * 70\)
Therefore \(n(n+1) = 69* 70\)
Even numbers from \(1\) to \(n = 2, 4, 6, 8 , 10 .....\)
\(2\) is the first even number.
\(n = 69 * 2 + 1 = 139\)
Answer (C)..._________________ Please Press "+1 Kudos" to appreciate.



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Re: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70
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20 Jul 2017, 01:06
==> If n=2m+1, the even numbers until 2m+1 becomes until 2m, and the sum of the even numbers is 2+4+…+2m=2(1+2+…m)=2m(m+1)/2=m(m+1). From m(m+1)=69*70, you get m=69, which becomes 2m+1=2(69)+1=139. The answer is C. Answer: C
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If the sum of all the even numbers from 1 to n(n: odd number) is 69*70
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02 Feb 2019, 15:17
sashiim20 wrote: MathRevolution wrote: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n?
A. 135 B. 137 C. 139 D. 141 E. 143 Sum of first \(n\) positive even numbers is given by formula \(= n(n+1)\)
Sum of all even numbers from \(1\) to \(n = 69 * 70\)
Therefore \(n(n+1) = 69* 70\)
Even numbers from \(1\) to \(n = 2, 4, 6, 8 , 10 .....\)
\(2\) is the first even number.
\(n = 69 * 2 + 1 = 139\)
Answer (C)..._________________ Please Press "+1 Kudos" to appreciate. Hello sashiim20 ! This was also my approach but I have a doubt, isn't n = 69? n(n+1) = sum of first evens = 69(69+1)Thank you so much in advance!



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Re: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70
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06 Feb 2019, 19:17
MathRevolution wrote: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n?
A. 135 B. 137 C. 139 D. 141 E. 143 My reasoning: Let's say the first number in our sequence is 2. However, we need to keep a mental note that we skipped the number 1 and to add it back in at the end. \(\frac{n}{2} * (2*2 + 2*(n1)) = 69*70\) Solve for n and we get n = 69. This value accounts for all the EVEN numbers in our sequence but we need to remember that for every even integer, we will have one odd integer. So 69 * 2 gives us 138 and since we took out the 1 at the start we need to add it back in to get 139 (answer choice C)



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If the sum of all the even numbers from 1 to n(n: odd number) is 69*70
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06 Feb 2019, 19:45
kchen1994 wrote: MathRevolution wrote: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n?
A. 135 B. 137 C. 139 D. 141 E. 143 My reasoning: Let's say the first number in our sequence is 2. However, we need to keep a mental note that we skipped the number 1 and to add it back in at the end. \(\frac{n}{2} * (2*2 + 2*(n1)) = 69*70\)
Solve for n and we get n = 69.This value accounts for all the EVEN numbers in our sequence but we need to remember that for every even integer, we will have one odd integer. So 69 * 2 gives us 138 and since we took out the 1 at the start we need to add it back in to get 139 (answer choice C) Hello kchen1994 ! How did you solve for n? \(\frac{n}{2} * (2*2 + 2*(n1)) = 69*70\) Solve for n and we get n = 69. I don't know what to do after it turns to be: (2)(69)(70) = 2n + 2n2 = 2n(1+n) (69)(70) = n(1+n) How do you know that 69 is n? Kind regards!



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Re: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70
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06 Feb 2019, 19:58
MathRevolution wrote: ==> If n=2m+1, the even numbers until 2m+1 becomes until 2m, and the sum of the even numbers is 2+4+…+2m=2(1+2+…m)=2m(m+1)/2=m(m+1). From m(m+1)=69*70, you get m=69, which becomes 2m+1=2(69)+1=139.
The answer is C. Answer: C Hello MathRevolution ! How do we know that n=2m+1? Kind regards!



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Re: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70
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06 Feb 2019, 20:03
MathRevolution wrote: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n?
A. 135 B. 137 C. 139 D. 141 E. 143 So yes, what did i do here. We have the sum as 69 * 70, this means in the formula for \(S_n\) = n/2 (FT + LT) n could have been divisible by 2 or FL + LT could have been divisible by 2 The only condition satisfying here was the former 69*2= 138 Key word : n: odd numberit will be the next term 139 Answer C
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Re: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70
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06 Feb 2019, 20:12
jfranciscocuencag wrote: kchen1994 wrote: MathRevolution wrote: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70, what is the value of n?
A. 135 B. 137 C. 139 D. 141 E. 143 My reasoning: Let's say the first number in our sequence is 2. However, we need to keep a mental note that we skipped the number 1 and to add it back in at the end. \(\frac{n}{2} * (2*2 + 2*(n1)) = 69*70\)
Solve for n and we get n = 69.This value accounts for all the EVEN numbers in our sequence but we need to remember that for every even integer, we will have one odd integer. So 69 * 2 gives us 138 and since we took out the 1 at the start we need to add it back in to get 139 (answer choice C) Hello kchen1994 ! How did you solve for n? \(\frac{n}{2} * (2*2 + 2*(n1)) = 69*70\) Solve for n and we get n = 69. I don't know what to do after it turns to be: (2)(69)(70) = 2n + 2n2 = 2n(1+n) (69)(70) = n(1+n) How do you know that 69 is n? Kind regards! So if you multiply n in we get \(69(70) = n + n^2\) Since 69*70 has already been factored for you you can solve this quadratic equation as so: \((n+70)(n69) = 0\) A series can never be negative so n must be 69. Keep in mind this only accounts for all the even values in our series. We must include all the odd values so 69*2. But since we skipped 1 when we started our series, we need to add it back in. Does this make sense?




Re: If the sum of all the even numbers from 1 to n(n: odd number) is 69*70
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