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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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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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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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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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