Tricky one, anybody

[tnet]

The arithmetic sequence a, a+d, a+2d, a+3d, ...,a+(n-1)d has the

following properties.

 

When the first, third, fifth, and so on are added to and including the

last term, the sum is 320

 

When the first, fourth, seventh and so on, terms are added, up to and

including the last term, the sum is 224

 

What is the sum of the whole sequence?

[mezarashi]

Interesting question. Well I guess this reinforces the understanding of how the summation formula for arithmetic series works. Remember the story about how Gauss added 1 to 100 in 5 seconds for his teacher although it was meant as punishment.

 

So we have an arithmetic series:

 

a1 + a2 + a3 + a4 + a5 + ... an

 

Normally, you would add this through the formula Sn = (n/2)(2a+(n-1)d) or more simply Sn = (n/2)(a1 + an)

 

which is basically how Gauss did his consecutive integer calculations. You add the last term to the first and multiply by the total number of terms divide by 2.

 

1.Given the conditions you stated, the first condition:

a1 + a3 + a5 + a7 +... an

 

You can see that the summation here will be

Sn1 = ((n+1)/4)(a1 + an) = 320

If this isn't clear. Write it out, perhaps to n=9. Write out all a1 to a9. You won't be adding all numbers this time, but how would you find the sum quickly? You still continue to add (a1 + a9), (a3 + a7) + a5. Note that (a1+a9)=(a3+a7). Instead of (n/2), we have here (n+1)/4.

 

2. Doing similar analysis for the second condition, we would get

Sn2 = ((n-2)/4)(a1 + an) = 224

 

Just assume that the last term will be part of the series, (i.e. if you have n=9, you will notice that the formula doesn't work, but n=10 will) In anycase, this fraction will cancel out later on.

 

So we have 2 equations Sn1 and Sn2. Let's group (a1 + an) together as a constant and let it be x. We then have two equations to solve with 2 unknowns. Solve for n. When you have n, you can then solve for (a1 + an).

 

Knowing both n and (a1+an), we know that the proper formula for the entire summation is:

 

Sn = (n/2)(a1+an)

 

as I mentioned, so plug in the numbers and you have your answer.

 

[Additional]

Further calculations for the values of a1 and d can be made as well, now that you know n. Just expand it accordingly and equivalate. I've put my answer below as a series that complies to all the conditions. Highlight.

 

 

 

 

192 160 128 96 64 32 0 -32 -64

a1 = 192, d = -32, n = 9, S9 = 576