IIT JEE , AIEEE Forum - Mathematics , Algebra - A problem on progressions..................................

nano0101

If a,b,c and p are disticnt real numbers such that:

(a² + b²)p² - 2(ab+bc)p + (b² + c²) <= 0

then a,b,c are in what kind of progression? AP, GP or HP?How or why?

taruntanuj007

now a term is a square so its an upward parabola and the expressions is less than zero so clearly it has 2 real roots and the expressions indicates those points where the curve is bellow x axis between the 2 roots!!!

here D < 0 since here the roots are not there

so 4(ab + bc)<= 4(a² + b²) (b² + c²) //= bcoz expressions has <=

after cancelling terms we have

2ab²c <= (ac)² + b⁴

Dividing by ab²c

2 <= ac/b²+ b²/ac

thus t + 1/t >= 2

(t - 1/t)²>= 0 which is always true but t is not zero

t >= 1/t

t² -1 >= 0 (t-1)(1+1) >= 0

t <= -1 and t >= 1

for t = 1 we have a GP here 's ur answr!!!!

Rate me if i am correct

nano0101

yes u r right thank you !

raghu_21

nice prob.....fin the discriminant...den find roots for p...the u will see

vishak_great

this has a simple solution

the equation can be written as

(ap-b)^2+(bp-c)^2<=0

so only equality holds (perfect square =>0)

hence

ap=b & bp=c

therefore p=b/a=c/b=r

so they arein gp

vishak

Avinash_Bhat

(a² + b²)p² - 2(ab + bc)p + (b² + c²) <= 0

(a²p² - 2abp + b²) + (b²p² - 2bcp + c²) <= 0

(ap - b)² + (bp - c)² <= 0

SO : (ap - b) = 0 and (bp - c) = 0

=> p = b/a and p = c/b

a/b = b/c => a,b,c in G.P.

iitkgp_bipin

(a²+b²)p² - 2(ab+bc)p + (b²+c²) <= 0

(ap-b)² + (bp-c)² <= 0

Since LHS >= 0

Hence (ap-b)² + (bp-c)² = 0

ap-b=0 and bp-c=0

p = a/b = b/c

Hence a,b,c are in GP.

nano0101

Thanks allllllllll of you for makin it much simpler !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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