IIT JEE , AIEEE Forum - Mathematics , Algebra

rohitarura

1.In an increasing GP,the sum of the first and the last term is 66,the product of the second and the last but one term is 128,and the sum of all the terms is 126.How many terms are there in the progresssion?? Ans-6

2.Sum of the series 1.1!+2.2!+3.3!+4.4!+.... upto n terms is?? Ans-(n+1)!-1

3.If n is a positive integer,prove that

4.Prove that ,where a,b,c are positive real nos.

5.Prove that ,where a,b,c are positive real nos.

6.Prove:

7.Prove:

sachinguptaiit

sachinguptaiit

konichiwa2x

6.

Bring the first term on the RHS to the LHS. So we need to prove

On simplification term by term,

konichiwa2x

4. BY AM-GM,

------ (2)

Given that ----- (3)

from (2) and (3)

konichiwa2x

3. I think you've made a typo. Did you mean?

If so, consider an GP with first term = 1 and common difference 2.

BY AM-GM inequality,

hsbhatt

4. An alternative is to use Rearrangement Inequality

hsbhatt

5. $\frac{ab}{c^3}+\frac{bc}{a^3} + \frac{ca}b^3} = abc \left[\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4} \right]$

So we have to prove that $\left[\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4} \right] \ge \frac{1}{abc} \left[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right]$

Putting a = 1/x b = 1/y and c = 1/z you have the previous inequality

rohitarura

Bhatt Sir,

Can you please explain me the fourth problem please

rohitarura

Abhijith Sir,or Bhatt Sir, please explain 4th problem again

spideyunlimited

Abhijit used 'a' number of a^3 s, 'b' number of b^3 s and 'c' number of c^3s
So their Arithmetic mean becomes a.a^3 + b.b^3 + c.c^3 / a + b + c

And GM becomes (a^3.b^3.c^3)^(1/a+b+c)
then the AM Gm inequality.

spideyunlimited

1.
Let the GP be : a, ar, ar^2 ....a.r^n

a + a.r^n = 66 = a(1 + r^n)
ar.a.r^(n-1) = 128 = a^2.r^n
So 66 = a (1 + 128/a^2)
66a = a^2 + 128
a^2 - 66a + 128 = 0
a = 66 +- rt(66^2 - 4(128)) / 2
a = 66 +- 62 / 2
a = 2, 64
Since, 66 = a(1 + r^n)
66 = 2(1 + r^n) OR 66 = 64 (1 + r^n)
Since r>1 (increasing GP) so we take
66 = 2(1 + r^n)
r^n = 32

And,

a(r^(n+1) - 1) / (r - 1) = 126
2(32r - 1) / (r-1) = 126
63r - 63 = 32r - 1
31r = 62
r = 2
and r^n = 32
2^n = 32
so n = 5

hsbhatt

Koni has already done the problem using AM-GM. But, there is a very handy inequality known as Rearrangement Inequality by which this becomes a one-liner

Rearrangement Inequality: You can read more on wikipedia or some other internet source, but the main idea is this - Suppose for example (x₁,x₂,x₃), (y₁,y₂,y₃,) (z₁,z₂,z₃) are ordered sequences, then you can assemble several sums taking numbers three at a time. One such product would be x₁y₂z₃+x₂y₁z₃+x₃y₃z₁

Rearrangement Inequality says that of all such sums the maximum is x₁y₁z₁+x₂y₂z₂+x₃y₃z_3.

This can be extended to any number of sequences with any number of terms

So now if you consider the repeated sequences (a,b,c) (a,b,c) (a,b,c) and (a,b,c)

then a⁴+b⁴+c⁴ >= a²bc+b²ac+c²ab = abc(a+b+c)

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