IIT JEE , AIEEE Forum - Mathematics , Algebra - $$$$$$$$$ KILLER QUESTION

Avinash_Bhat

An = ( 1.2.3 + 2.3.4 + 3.4.5 + .............................. upto n terms )

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n( 1.2 + 2.3 + 3.4 + ................................... upto n terms )

_{[ n ][ ]} An = x / y ; where x and y are integers and x / y is in its lowest form.

If (y - x)p³ + y^1/2r(pqx - (x + 1)r²) = (x - y)q³then which of the following is/are true ? (p, q, r are distinct).

a) p, q, r in A.P.

b) p, q, r in H.P.

c) p + qw - 2rw²= 0

d) p + qw² + 2rw = 0

e) none of these ; w is the complex cube-root of unity.

catch_arnnie

in the 2nd question, the 1st option is not clear...

Avinash_Bhat

HEY .................................... !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

IT IS A SINGLE QUESTION AND ............. THE VALUES OF 'x' AND 'y' ARE USED IN THE EQUATION

Avinash_Bhat

IS IT THAT DIFFICULT ??????????????

coolguy2007

I have solved to an extent where the answer could be either a) or b). What was the answer given in the book(or whatever material)?

vishak_great

please check up

from first relation we have 2b=a+c

from second relation we have c^2 = bd

from third relation we have c^2=ae

log_a(c^2) =log_ea+log_ce

log_a ae = log_ea + 1/log_ec

log_ec+log_ac=log_eclog_ea+1

putting c= ^[2]ae and simplifying

we get

(log_ea)^{3 _{= 1}}

_{^{which further means a= b=c=d=e}}

_{^{as a,b,c,d,e, are real}}

_{^{(which is against your first claim)}}

and x=0 and p^{3 _{+ q}3 _{+ r}3 _{= 0}}

please find error if any in the solution

Avinash_Bhat

Hey VISHAK,

HOW DID YOU GET :

log _a ae = log _e a + 1 / log _e c

=> log _e c + log_a c = (log _e c)(log_e a) + 1

Avinash_Bhat

HEY, .....................................

NO ONE WITH AN ANSWER ?????????

vishak_great

log_ea, log_ac, log_ce are in A.P.

so log_ac^2 = log_ea+log_ce

ramyadiamond

I'm also getting a=b=c=d=e, i think vishak is right.

Avinash_Bhat

I THINK YOU ARE RIGHT................................

NOW TRY THE EDITED QUESTION.

ramyadiamond

After simplifying the expression, i'm getting

An= 3(n-2)/4n

and the limit gives me x/y=3/4

and in the given question, i could modify it in terms of y as the quadratic expression of y^1/2 ., so if u take y^1/2=M

then u get an expression as

M²(3pqr-3r⁴) + (p³+q³)M -4r⁴=0

but i am unable to continue from here

vishak_great

numerator = _{[ ]}^{[ ]} n(n+1)(n+2) =[n(n+1)(n+2)(n+3)]/4 (use method of differences

or summation formula)

denominator = n(_{[ ]}^{[ ]} n(n+1)) = [n(n+1)(n+2)](n)/3

An = [3(n+3)]/4n= (3/4) (1+ 3/n)

as n tends to infinite An =3/4 x=3,y=4

substituting these values we have

p^3 + q^3 = 8(r)^3 - 6pqr

p^3 + q^3 + (-2r)^3 = 3(p)(q)(-2r)

so p+q - 2r = 0 p,r,q are in a.p

or p = q = -2r p+q+4r = 0

or both then p=q=r=0

you have not given conditions for p,q,r Please check up

vishak

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