Arithmetically Progressive
If A , B , C , D are consecutive terms in an arithmetic progression, what is the value of
C 2 − B 2 D 2 − A 2 ?
Assume C 2 − B 2 = 0 .
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6 solutions
how can we substitute in Arithmetic progression ? I can't get it clearly, and How may I assign these substituted value, I have no idea about it, give me some idea about it.
What is a non-constant aarithmetic progresion?
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The constant arithmetic progression is c , c , c , c , … , i.e. where the common difference is 0.
The reason for adding in "non-constant", is so that the denominator is not 0.
I rephrased the question.
Did it the same way except I substituted B = A + d , C = A + 2 d , D = A + 3 d
It's quite surprising to know this !
lets say A=a-2d , B=a-d , C=a+d , D=c+2d . now simply putting these values on the given expression we get 2 which is not in the option. so i gives there is problem. please recheck it.
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Note that a − 2 d , a − d , a + d , a + 2 d is not an arithmetic progression. The differences are d , 2 d , d .
Let us assume that A , B , C and D are the terms of an arithmetic progression with common difference d .
Now we have the given expression C 2 − B 2 D 2 − A 2 = ( C + B ) ( C − B ) ( D + A ) ( D − A ) .
And since we know that D + A = C + B because sum of the terms of an AP equidistant from the beginning and the end is always constant irrespective of the distance, therefore further we have the given expression = ( C − B ) ( D − A ) = d 3 d = 3 . □
4 terms are a-d, a, a+d, a+2d hence it reduces to (2a+d)3d/(2a+d)d = 3
I think the simplest way of solving this is to just stick numbers in.
A =1
B =2
C =3
D = 4
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
16 - 1 / 9 - 4
15/5 = 3
So, a solution is 3. However, there's a possible answer that says "It depends". That means we need to repeat this to see if we get 3 again. We'll keep it simple. Just use the next set of numbers
A =2 _ _ 2^2 = 4 B =3 _ 3^2 = 9 C =4 _ 4^2 = 16 D = 5 _ _ 5^2 = 25
25 - 4 / 16 - 9
= 3 So, 3 is our answer.
It's even easier if you sub a=-1 b = 0 c=1 and d=2
@Adnan Ilyas :Two examples isn't a concrete proof in mathematical terms - all it means is that you haven't found an exception yet - and considering all the sets of 4 numbers you could use - 2 examples is a very small sample set. To proove these things conclusively - you need to use algebraic manipulation - i.e. proove that no matter what number set you use - yet get the same answer - and you can only do that with algebra (or in this case you can use logic - as suggested by Sandeep above).
Unsure whether this constitutes formal proof, because it's a particular case, though it seems obvious that its result would apply to any arithmetic progression.
Susbitute D for the expression (x+3), and A for the expression (x), with B and C covering the intermittent arithmetic integer values (x+1) and (x+2) respectively.
Expanding the brackets and collecting like terms leaves us with (6x+9)/(2x+3) = 3
I found this identify surprising when I first saw it.
Simply substitute A = a , B = a + d , C = a + 2 d , D = a + 3 d to obtain
C 2 − B 2 D 2 − A 2 = ( C − B ) ( C + B ) ( D − A ) ( D + A ) = d ( 2 a + 3 d ) 3 d ( 2 a + 3 d ) = 3 .