I have a very tough problem in AP. Even my teacher has no idea what to do?

Question

The problem is long and tough to type, so please bear with me. I hope the message gets across:

[1/(a1*an)] + [1/(a2*an - 1)] + ... + [1/(an * a1)] ..... (1)

[2(a1 + an)][(1/a1) + (1/a2) + (1/a3) ... + (1/an)] ..... (2)

If a1, a2, a3, a4 .... , an -1, an are in AP, prove that (1) = (2)

If the proof is tough to type or too long, mail me here:
[e-mail]

The dots are there in the question.

The question is NOT mistyped.

a is the term. n, n - 1, 1, 2 are all subscripts. Write an , an - 1 the way you would write O2 for oxygen. a1, a2, a3, ... an - 1, an are terms of an AP. I repeat, the question is NOT mistyped. The first and last terms, second and second last terms and so on ARE the same. What nags me is the proof.

Horatio, I'm surprised that you don't know that AP stands for Arithmetic Progression

I am SORRY, guys.

The question was mistyped.

(2) is [2/(a1 + an)][(1/a1) + ... ]
NOT 2(a1 + an)

VERY VERY SORRY!!!

Answer 1

What is AP? It better be something really special, because your formula is already wrong when n = 4:

(1)
2/a1a4 + 2/a2a3 = 2(a2a3 + a1a4)/a1a2a3a4

(2)
2(a1 + a4)(1/a1 + 1/a2 + 1/a3 + 1/a4)
2(a1 + a4)(a2a3a4 + a1a3a4 + a1a2a4 + a1a2a3)/a1a2a3a4

You should immediately see that these aren't equal, but just in case you don't, set a1 = a2 = a3 = a4 = 1. Then (1) = 4 and (2) = 16.

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Did you even read what I wrote? I had no problem understanding that a1...an are different terms of some kind of sequence. My point is that unless you define that sequence very carefully, there is no way this equality is true. Go ahead and try it yourself. See which values of a1..an it actually works for. I bet that it doesn't work for almost any set of a1..an you come up with.

P.S. Mayur, what the hell is that? You are just copying the first answer you see, even if that first answer was completely useless?

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I don't think AP = arithmetic progression is a standard abbreviation, but I could see how it would become custom in your class to make such an abbreviation. Anyway, armed with that knowledge and your fixing of the problem statement, the problem becomes pretty easy.

Consider:

1/a1 + 1/an = (a1 + an)/a1an

So therefore we can decompose the first term as follows:

1/a1an = 1/a1(a1 + an) + 1/an(a1+an)

For the second term, we have:

1/a2an-1 = 1/a2(a2 + an-1) + 1/an-1(a2 + an-1)

But, because we have an arithmetic progression, a2 + an-1 = a1 + an! So we can actually write the second term as

1/a2(a1+an) + 1/an-1(a1+an)

Do the same thing for each of the terms, and you get

2(1/a1(a1+an) + 1/a2(a1+an) + 1/a3(a1+an) ... 1/an(a1+an))

Factor out the 1/(a1+an) and you get equation (2) exactly.

P.S. Now you have something intelligent to plagiarize, Mayur. Knock yourself out.

P.P.S. This proof does work if n is odd as well.

Answer 2

Assuming a1,... , an is in AP ;) typo I think.
And also n is even, otherwise it won't work ;)

You should have atleast verified on an AP.

Simple problem :) shame on your teacher ;)


Starting from (2)....

Pair i-th term from beginning and ending..
i.e. (1/a_1 + 1/a_n) + (1/a_2+1/a_(n-1)) + (1/a_3+1/(a_(n-2)) +...

Numerator of all terms
= ai+a(n-i+1) = a+(i-1)d + a+(n-i)d = 2a + (n-1)d
Denominator = a_(i) * a_(n-i+1)

Numerator independent of i, hence = a1+an, take it out :)

Canceling Q.E.D.

Answer 3

did you mess up? the first and last terns in (1) are the same is it supposed to be 1/(an*an-1)?

Answer 4

i think there is a flaw in ur question.
ur expression 2 must be [2/(a1+an)[...]

first note that if the AP is an= a+(n-1)d
then, an+a1=a(n-1)+a2=...=2a+(n-1)d



exp1
= [1/a1*an]+.....+[1/an*a1]
=[1/(a1+an)][(a1+an)/(a1*an)+(a1+an)/(a2*an-1)+...+(a1+an)/(a1*an)]
=[1/(a1+an)][(a1+an)/(a1*an)+(a2+an-1)/(a2*an-1)+...+(an+a1)/(a1*an)]
=[1/(a1+an)][{(1/a1)+(1/an)}+{(1/a2)+(1/an-1)}+......]
=[1/(a1+an)][2{(1/a1)+(1/a2)+...+(1/an)}]
=[2/(a1+an)][(1/a1) + (1/a2) + (1/a3) ... + (1/an)] which is my expr.2
and not urs. cos if they are equal, then
a1+an=1/(a1+an)
a1+an=1, which is not true for all APs

Answer 5

I think you have too many spaces in what you typed in. That's why the ellipses appear.

Answer 6

did you mess up? the first and last terns in (1) are the same is it supposed to be 1/(an*an-1)?