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( c 2 a 2 + b 2 ) ⎣ ⎡ ( a b c a 3 + b 3 + c 3 ) 6 − 2 9 ⎦ ⎤
Consider a polynomial such that its roots are in arithmetic progression and the squares of these roots are also in an arithmetic progression in the same order.
If the roots are a , b , c such that a , b , c = 0 , find the value of the said expression above.
The answer is 1400.
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4 solutions
Awesome question and excellent explanation.
Let me post my bashy way in the comments just to show a different non-elegant approach.
My method is identical to this one till the part where he proves b 2 = a c implying that a , b , c forms a geometric progression. As such, we use the general form of GP to reduce the expression (noting the homogeneity of the expression and that a , b , c = 0 ). Take r as the common ratio of the GP sequence { a , b , c } and the required expression as E . We get,
E = r 4 1 + r 2 ( r 2 4 ( 1 + r 3 + r 6 ) 6 − 2 9 )
Now, from the fact that both the sequences ( a , b , c ) and ( a 2 , b 2 , c 2 ) are ordered AP sequences, we get two equations:
{ 1 + r 2 = 2 r 2 r 2 = 1 + r 4 ⟹ r + r 3 = 2 r 2 = 1 + r 4 ⟹ r 4 − r 3 − r + 1 = 0
This polynomial equation has only one real solution, which is r = 1 . Now, simply substitute this value of r into the reduced expression for E and you get,
E = 1 1 + 1 ( 1 3 6 − 2 9 ) = 1 4 0 0
Note: By ordered AP sequence, I'm referring to the fact that the order of the AP sequence is as written there and cannot be permuted randomly. It can only be permuted in at most another way: reversely. (last and first elements switch)
How about ( a , b , c ) = ( 7 , 1 , − 5 ) ?
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( a 2 , b 2 , c 2 ) = ( 4 9 , 1 , 2 5 ) ⟹ a 2 , c 2 , b 2 are in GP ⇒ a 2 , b 2 , c 2 are in GP
⇒ denotes "Not implies"
I agree that the problem needs to be rephrased a bit to mention the ordering.
a 2 , b 2 , c 2 is not in A.P.
OMG Dude! This didn't strike me! Good one indeed. Perfect. "One of the finest minds in the country."
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Thanks hawkeye .. xD
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This was quite an easy problem.
Cool method.but after knowing that a,b,c are in AP as well as in GP we can write a=b=c. So substituting in eqn we get the ans 1400
i took a, c, b in ap
We know: b-a = c - b
and
b²-a² = c²-b² ->
-> (b-a)(b+a) = (c-b)(c+b) ->
-> a+b = c+b ->
a = b = c
The expression is equal to 2. (729-9) = 1400
Dividing b − a in the equation ( b − a ) ( b + a ) = ( c − b ) ( c + b ) ⟹ b = a , resulting in a contradiction.
Way too easy (though I didn't get it at first immediately). assuming that the roots are a,a+d,a+2d and using the fact that the squares of the roots are also in AP in the same order, we can write ( a + d ) 2 − a 2 = ( a + 2 d ) 2 − ( a + d ) 2 . Simplifying the said expression gives d 2 = 3 d 2 , which means that d=0. Thus the 3 roots are equal. Game over :)
A one second problem.take the 3 roots as 1,1,1... 1,1,1 are in AP Even their squares are in AP Substitute and get the ans as 1400
According to the given condition ,
a + c = 2 b … ( 1 ) a 2 + c 2 = 2 b 2 … ( 2 ) ⇒ ( a + c ) 2 − 2 a c = 2 b 2 ⇒ ( 2 b ) 2 − 2 a c = 2 b 2 … from (1) ⇒ 2 a c = 2 b 2 ⇒ b 2 = a c … ( 3 )
Note: We get an astonishing result that a , b , c also form a geometric progression!
Substituting ( 3 ) in ( 2 ) ,
a 2 + c 2 = 2 a c ⇒ ( a − c ) 2 = 0 ⇒ a = c … ( 4 )
Substituting ( 4 ) in ( 3 ) ,
b 2 = a 2 ⇒ a = b ⇒ a = b = c (by transitivity)
This makes the given expression:
( a 2 2 a 2 ) ( ( a 3 3 a 3 ) 6 − 2 9 ) = ( 2 ) ( 3 6 − 2 9 ) = 2 ( 7 2 9 − 2 9 ) = 2 ( 7 0 0 ) = 1 4 0 0