# Questions of geometric progression.

sankalpmittal

## Homework Statement

1. Prove that 1111111......65 times is not a prime number.

2. If a,b,c,d are in GP , then (a2+b2+c2)(b2+c2+d2) is :

(a) (ab+ac+bc)2
(c) (ab+bc+cd)2
(d) None of the above

## Homework Equations

Formulas of GP :

1.Sn = a(1-rn)/(1-r)
2.a,b,c in GP , then b2=ac
3.a,b,c,d,e,....,z in GP , then
az=by=cx=dw =.....

4.Tn = arn-1

## The Attempt at a Solution

1. The number can be written as

N = 1+10 + 102 + 103 +.... + 1064
N= (1065-1)/9

Now how to proceed ?

2. Well I tried using b2=ac but to no avail.

Homework Helper
http://en.wikipedia.org/wiki/Primality_test :)
Though I notice that 1/9 = 0.11111111...

b2=ac only applies if the GP has three terms {a,b,c}

if it has an odd number of terms then one of the products will be with itself.

the general rule is:
if $$a_1, a_2,\cdots,a_{n-1},a_n$$ is a GP, then $$a_1 a_n = a_2 a_{n-1} = a_3 a_{n-2} = \cdots$$

So for just the four ... what does this become?

Homework Helper

## Homework Statement

1. Prove that 1111111......65 times is not a prime number.

## The Attempt at a Solution

1. The number can be written as

N = 1+10 + 102 + 103 +.... + 1064
N= (1065-1)/9

Now how to proceed ?

65=5˙13, so the number can be also written as N= ((105)13-1)/9. Can you factor it?

ehild

sankalpmittal
http://en.wikipedia.org/wiki/Primality_test :)
Though I notice that 1/9 = 0.11111111...

b2=ac only applies if the GP has three terms {a,b,c}

if it has an odd number of terms then one of the products will be with itself.

the general rule is:
if $$a_1, a_2,\cdots,a_{n-1},a_n$$ is a GP, then $$a_1 a_n = a_2 a_{n-1} = a_3 a_{n-2} = \cdots$$

So for just the four ... what does this become?

Ok , first stick to first question. And I've gotta use GP to prove that 11111.....65 times isn't a prime number.

65=5˙13, so the number can be also written as N= ((105)13-1)/9. Can you factor it?

ehild

Ah ! Do you mean using an-bn formula ? Or you mean doing plain factorization ?

I apologize for such a delay. Things were slightly downhill for me.

Homework Helper
Ah ! Do you mean using an-bn formula ? Or you mean doing plain factorization ?

Yes, you always can factor out (a-b) from an-bn,
1065-1=((105))13-1,
a=105 b=1.

ehild

sankalpmittal
Yes, you always can factor out (a-b) from an-bn,
1065-1=((105))13-1,
a=105 b=1.

ehild

Woops !

I forgot the formula for an-bn. Ok , so I'm doing simple factorization..

N= ((1013)5-1)/(10-1)

Now factorizing 1065-1 by 1013-1 , I get : 1052 + 1039 + 1026 + 1013 + 1

So ,N= (1052 + 1039 + 1026 + 1013 + 1)(1013-1)/(10-1)

On further simplifying , I get :

N=(1012 +1011 + 1010 + 109 + 108 + 107 + .... + 1)(1052 + 1039 + 1026 + 1013 + 1)

Thus N or given number is a multiplication of two non zero numbers. So it has to be composite number. Proved!!

Thanks ehild and Simon !!

(Wait , Am I methodically correct in question 1 ?)

All right now to second problem :

"If a,b,c,d are in GP , then (a^2+b^2+c^2)(b^2+c^2+d^2) is :

(a) (ab+ac+bc)^2
(c) (ab+bc+cd)^2
(d) None of the above"

So we have

Now how to proceed further ?

Last edited:
Homework Helper
Woops !

I forgot the formula for an-bn. Ok , so I'm doing simple factorization..

N= ((1013)5-1)/(10-1)

Thus N or given number is a multiplication of two non zero numbers. So it has to be composite number. Proved!!

It is OK, but I suggested the easier way:
$$N=\frac{10^{65}-1}{9}=\frac{(10^{5})^{13}-1}{9}=\frac{99999(10^{60}+10^{55}+...+1)}{9}=11111(10^{60}+10^{55}+...+1)$$

All right now to second problem :

"If a,b,c,d are in GP , then (a^2+b^2+c^2)(b^2+c^2+d^2) is :

(a) (ab+ac+bc)^2
(c) (ab+bc+cd)^2
(d) None of the above"

So we have

Now how to proceed further ?

I would do it with "brute force" using that b=ar, c=ar2, d=ar3, and write up the expressions in terms of a and r.

ehild

sankalpmittal
I would do it with "brute force" using that b=ar, c=ar2, d=ar3, and write up the expressions in terms of a and r.

ehild

That is going to take a lot of time , right ? Does it not ? And that is making use of objective problem. What if it were not an objective problem. (Oops I just now understood that such problem can only be given as objective problem ! Sorry !!)

Ok , but by so called "brute force" , I've to first find (a^2+b^2+c^2)(b^2+c^2+d^2) in terms of a and r. Then I've to solve for all the options also , in terms of a and r. Any other method ?

By brute force , I have :

(a2+b2+c2)(b2+c2+d2) = a2(1+r2+r4) a2r2(1+r2+r4) = a4r2(1+r2+r4)2

All right , now I'll have to do all this for all the options...

Well , I'm on it...

(a) (ab+ac+bc)2 =( a2r + a2r2 + a2r3)2 = a4r2(1+r+r2)2

So option a is out...

I can simplify it to ,

(a2r2+a2r5+ar3)2

a2r4(1+r3+r)2

b is out.

(c)(ab+bc+cd)2

(a2r+a2r3+a2r5)2

a4r2(1+r2+r4)2

So (c) is correct option. Yes !! Thanks ehild !!

But do you know any other method ?

Homework Helper
So (c) is correct option. Yes !! Thanks ehild !!

But do you know any other method ?

You are unsatiable Was not it an easy and short solution?

ehild

sankalpmittal
You are unsatiable Was not it an easy and short solution?

ehild

Yes , it was a short solution though. I've heard that (a2+b2+c2)(b2+c2+d2) = (ab+bc+cd)2 is an identity also. Lagrange's identity you call it ? Or is it Cauchy Schwarz inequality - as I can see here :

Thanks ! Once again for help !!

Homework Helper
Yes , it was a short solution though. I've heard that (a2+b2+c2)(b2+c2+d2) = (ab+bc+cd)2 is an identity also. Lagrange's identity you call it ? Or is it Cauchy Schwarz inequality - as I can see here :

Thanks ! Once again for help !!

It is an inequality in general,

(a2+b2+c2)(b2+c2+d2) ≥ (ab+bc+cd)2.

I used to be a bad student, I did not like to memorize theorems, and if I did , I could not remember the name. I preferred to prove a theorem myself. It was fun, memorizing was difficult and boring :tongue2:

Anyway, thank you for the file, I saved it and try to remember it...

ehild

sankalpmittal
It is an inequality in general,

(a2+b2+c2)(b2+c2+d2) ≥ (ab+bc+cd)2.

I used to be a bad student, I did not like to memorize theorems, and if I did , I could not remember the name. I preferred to prove a theorem myself. It was fun, memorizing was difficult and boring :tongue2:

Anyway, thank you for the file, I saved it and try to remember it...

ehild

I know proving is more fun , but I am preparing for engineering entrance exams. There , I have to do each question in 1 minute. So , my guess is that its better to use analytical skills to derive shortcut formulas by yourself and then memorize them. But I never get satisfied by a formula , if I do not know its proof.

That's Ok.... But what I have doubt still is regarding :

It is OK, but I suggested the easier way:
$$N=\frac{10^{65}-1}{9}=\frac{(10^{5})^{13}-1}{9}=\frac{99999(10^{60}+10^{55}+...+1)}{9}=11111(10^{60}+10^{55}+...+1)$$

What did you take factor out of what ?

I would do it with "brute force" using that b=ar, c=ar2, d=ar3, and write up the expressions in terms of a and r.

ehild

If I take 105=x , say and then

((105)13-1)/(10-1)

N = (x13-1)/9
N= (x-1)(x12+x11+x10....+1)/9
N= (1+10+102+103+104)(1060+1055 +.....1)

Oh !! This is what you've got !! How did you do this in just one line !?

Thanks again for your endeavors , ehild !!

Homework Helper
I know proving is more fun , but I am preparing for engineering entrance exams. There , I have to do each question in 1 minute. So , my guess is that its better to use analytical skills to derive shortcut formulas by yourself and then memorize them. But I never get satisfied by a formula , if I do not know its proof.
It is very good. Never believe anything without understanding the proof or proving it yourself.

If I take 105=x , say and then

((105)13-1)/(10-1)
N = (x13-1)/9

Do not forget that 105=100000 (5 zeros), and 100000-1=99999.
N= (x-1)(x12+x11+x10....+1)/9=99999(......)/9=11111(....) (as 99999 divided by 9 is 11111....) Never forget what you learnt in the first or second class :tongue2:.

ehild