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Algebra

Blazing goIITian

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19 Dec 2007 23:37:33 IST

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jee 1998 progression queston

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Let x be an arithmetic mean and y,z are two geometric means b/w any two positive nos then y3+z3/xyz =??

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Himanshu

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Joined: 19 Feb 2007

Posts: 3834

20 Dec 2007 01:17:39 IST

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let 2 no.s be a nd b

x = a + b / 2 , 2x = a + b

let

a , y , z , b are in G.P

so common ratio is r

y = ar , z = ar² , b = ar³

y³ + z³ = (ar)³ + (ar²)³

= a³r³ + a³r⁶...............1)

as b = ar³........put r³ = b/a

1) becomes

= a³(b/a) + a³ (b/a)²

a²b + ab² = ab ( a + b).....

as we know....a + b = 2x (x is A.M.)

so it becomes ab (2x) = 2x.ab

yaar m done till here....ye ab kaise jayega...ye ni ho raha

ans i 2 only na??

see the next post

Indzee ...

Hot goIITian

Joined: 2 Dec 2007

Posts: 124

20 Dec 2007 01:27:40 IST

Like 1 people liked this

Got the answer from Himanshus post,

Let a and b be the two positive nos,

Given,

x = (a + b) / 2 = AM ..(i)

a, y, z, b are in G.P

To find: (y³ + z³) / xyz

Let the common ratio be r,

Therefore,

y = ar;

z = ar²;

b = ar³;

Therefore r³ = b/a ..(ii)

y³ + z³ = a³r³ + a³r⁶ ..(iii)

Now, Substitute (ii) in (iii), we have,

y³ + z³ = a²b + ab² = ab (a + b) ..(iv)

yz = (ar)(ar²) = a²r³ = ab (because b = ar³) ..(v)

Substituting the values of equation (i), (iv) and (v) in

(y³ + z³) / xyz

We have,

(y³ + z³) / xyz = ab (a + b) / ab (a + b) / 2

= 2

Is the answer right???

Himanshu

Forum Expert

Joined: 19 Feb 2007

Posts: 3834

20 Dec 2007 01:30:31 IST

Like 3 people liked this

let 2 no.s be a nd b

x = a + b / 2 , 2x = a + b

let

a , y , z , b are in G.P

so common ratio is r

y = ar , z = ar² , b = ar³

y³ + z³ = (ar)³ + (ar²)³

= a³r³ + a³r⁶...............1)

as b = ar³........put r³ = b/a

1) becomes

= a³(b/a) + a³ (b/a)²

a²b + ab² = ab ( a + b).....

yz = (b^1/3.a^2/3)(b^2/3.a^1/3) = ab

y³ + z³ = 2.x (yx).................2x = a+b

y³ + z³/xyz = 2

hope u got it

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