How to use the geometric progression of terms a,b,c,d in identity: (a-d)^2=(b-c)^2+(c-a)^2+(d-b)^2
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The terms a, b, c and d form a geometric progression. So we can write the terms as b = ar , c = ar^2 and d = ar^3
We have to prove (a - d)^2 = (b - c)^2 + (c - a)^2 + (d - b)^2
We start with the left hand side:
(a - d)^2
=> (a - ar^3)^2
=> a^2(1 - r^3)^2 ...(1)
The identity cannot be proved for (a - d)^2 = (b - c)^2 + (c - a)^2 + (d - b)^2, instead it should be (a - d)^2 = (b - c)^2 - (c - a)^2 + (d - b)^2
(b - c)^2 - (c - a)^2 + (d - b)^2
=> ( ar - ar^2)^2 - ( ar^2 - a)^2 + (ar^3 - ar)^2
=> a^2[(1 - r^2)^2 - (r^2 - 1)^2 + ( r^3 - 1)^2]
=> a^2[ 1 + r^4 - 2r^2 - r^4 - 1 + 2r^2 + r^6 + 1 - 2r^3]
=> a^2[ r^6 + 1 - 2r^3]
=> a^2( 1 - r^3 )^2 ...(2)
From (1) and (2) we get (a - d)^2 = (b - c)^2 - (c - a)^2 + (d - b)^2
The identity that can be proved using the terms a, b, c and d of a GP is (a - d)^2 = (b - c)^2 - (c - a)^2 + (d - b)^2
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Well, you can use the fact that each term of a geometric progression, beginning with the 2nd, is the geometric average of the neighbor terms.
We'll write this constraint mathematically:
b^2 = ac; c^2 = bd
Now, we'll expand the squares both sides:
a^2 - 2ad + d^2 = b^2 - 2bc + c^2 + c^2 - 2ac + a^2 + d^2 - 2bd + b^2
We'll eliminate a^2 and d^2 both sides:
-2ad = 2b^2 + 2c^2 - 2bc - 2ac - 2bd
But b^2 = ac; c^2 = bd=>
=> -2ad = 2b^2 + 2c^2 - 2bc - 2b^2 - 2c^2
-2ad = - 2bc
We'll divide by -2:
ad = bc
We'll divide by a and c:
b/a = d/c = q, the common ratio.
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