Arithmetic Progression

[Kodwo]

I have been struggling to understand the question and answer below I came across in my GMAT class.

Can anyone explain the Q and A below and set another question very similar to this question for me to try and resolve?

*Q1.The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?*

*A.* Sol: First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

[karan13]

hi let me try this....

Lets start by stating the formula.
Sn=n/2(2a+(n-1)d)
here sn is the sum of all terms, n is the no. of terms, a is the first term and d is your common difference between each term.

Since the common difference is 2 between each term(only even nos)this is an AP and therefore the above formula is applicable.

Now your n=n-1+1(read how to find the no. of terms this is a direct formula) which is equal to n.
d=2 as discussed above
a=2(note a is not 1 cos they have only asked for even nos and 1 can not be a part of that list ,therefore first term is 2)

Now put these values in the formula above:
79*80=n/2(2*2+(n-1)2)
take 2 common and simplify you get 79*80=n(n-1)
to solve this the method i used to calculate n is as follows.. (note that there is no need to multiply that big no 79*80 and waste ur time)Simplify the above you get quadratic eq
n2-n-(79*80)=0(note the factors to satisfy the equations are already given so easy to simplify)
n2-80n+79n-(79*80)=0
n(n-80)+79(n-80)=0 (you have ur factors now)
n=-79 and n=80
obviously n can not be negative therefore ans is 80

hope it helps...

[jnelson0612]

Kodwo, hoping this is making more sense to you.

[sadat.sathak]

@ Karan13

obviously n can not be negative therefore ans is 80

This is not the correct answer.
I think you overlooked a critical point in the question stem.

*Q1.The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?*

There is a simple formula for calculating sum of consecutive numbers from 1 to n-->n(n+1)/2 (refer to OG 12 PS 157)

Now since we are calculating in terms even number we need to account for that.
So the formula becomes--> 2n(n+1)/2, which is basically (2s cancel out) = n(n+1)=79*80 hence n=79 & n+1=80. And since n is given as odd, n=79. (in fact, I think it was given explicitly as odd in the question stem in order to avoid any confusion!)

Even by using the formula you have given-->Sn=n/2(2a+(n-1)d)
we could derive 79*80= n/2(2*2(n-1)2)= 79*80=n(n+1) (no need for messy quadratics)

Side note: If you can intuitively understand how this formula is derived and how it makes logical sense, you can gain a deeper understanding and a better grasp instead of blindly apply formulas
http://betterexplained.com/articles/techniques-for-adding-the-numbers-1-to-100/

[JohnHarris]

Are you sure that is the exact wording of the problem? At first look, I (and apparently the others here), read it as "The sum of the first n even numbers is 79*80, ..."
and the answers here reflect that. However, if you read the problem the way it is stated, you get a different answer.

To go the long way around, what the question is asking is: If
2 + 4 + 6 + ... + 2 [(n-1)/2 -1] + 2 [(n-1)/2] = 79 * 80
what is the value of n. That is since n is odd, n-1 is even and the sum is only of even number up n-1, i.e. between 1 and n.

Thus we have
SUM{[2 j], j = 1 to [(n-1)/2] }
For convenience let m = (n-1)/2. Then we can write
SUM{[ 2j], j = 1 to m } = 79 * 80
or, as before with m replacing n,
m (m+1) = 79*80
Since this is a quadratic equation there are only two solutions and, since 79*80 is positive, both m and m+1 must be positive or both must be negative. Since m negative is not allowed and m = 79 works, m=79 must be the answer. Since n = 2 m + 1 , n = 159.

For another question try "Let S be the sum of the first n numbers divisible by 3. The value of S is 198. What is n?" Hint: It helps if you can recognize that 198 is equal to (3/2) * 11 * 12

[tim]

Nice work, John..