Problem in Algebra and Geometric sequence

Question

I need help on this one question which is in Algebra and on Geometric progression. The question is as follows: In a geometric sequence prove that: $(b-c)^2 + (c-a)^2 + (d-b)^2 = (d-a)^2$.

Thanks, Sudeep

Answer 1

We need to eliminate $b,c$

we have $$\frac ba=\frac cb=\frac dc\implies b^2=ac,c^2=bd,bc=ad$$

$$(b-c)^2+(c-a)^2+(d-b)^2$$

$$=2b^2+2c^2-2bc-2ca-2bd+a^2+d^2$$

$$=2(b^2-ca)+2(c^2-bd)+a^2+d^2-2ad\text{ as }bc=ad$$

$$=a^2+d^2-2ad$$

Answer 2

Let $$\frac dc=\frac cb=\frac ba=k\implies b=ak,c=bk=ak^2,d=ck=ak^3$$

$$(b-c)^2+(c-a)^2+(d-b)^2=a^2\{(k-k^2)^2+(k^2-1)^2+(k^3-k)^2\}=a^2(k^6-2k^3+1)=\{a(1-k^3)\}^2=(a-d)^2$$