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

Question

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

generis · Accepted Answer

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 terms

The 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 terms

The 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

sashiim20 · Answer

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.

MathRevolution · Answer

==> 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

The answer is C.
Answer: C

jfranciscocuencag · Answer

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!

NinetyFour · Answer

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*(n-1)) = 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)

jfranciscocuencag · Answer

Hello kchen1994 !

How did you solve for n?

\frac{n}{2} * (2*2 + 2*(n-1)) = 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!

jfranciscocuencag · Answer

Hello MathRevolution !

How do we know that n=2m+1?

Kind regards!

KanishkM · Answer

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 number

it will be the next term 139

Answer C

NinetyFour · Answer

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)(n-69) = 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?

bumpbot · Answer

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

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