If a b c d are in g.p. then prove that (a+b)^2, (b+c)^2, (c+d)^2 are in g.p.
Answers
(a^2+b^2+c^2)(b^2+c^2+d^2)=a^2.b^2+b^2.c...
{ a.c=b^2; b.d=c^2 }..........(1) and (2)
dividing (1)/(2)
a.c/c^2=b^2/b.d
then a/c=b/d
ad=bc (squaring on both sides)
a^2.d^2=b^2.c^2
a.d.a.d=b.c.b.c (a.d=b.c)
abcd=b^2c^2 and also a^2.d^2=abcd.....(3) and (4)
a^2.c^2=a.c.b^2 and b^2.d^2=b.d.c^2.....(5) and (6)
b^4=b^2(a.c) and c^4=c^2(b.d)..............(7) and (8)
Explanation:
a, b, c, d are in G.P.
Therefore,
bc = ad … (1)
b2 = ac … (2)
c2 = bd … (3)
It has to be proved that,
(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2
R.H.S.
= (ab + bc + cd)2
= (ab + ad + cd)2 [Using (1)]
= [ab + d (a + c)]2
= a2b2 + 2abd (a + c) + d2 (a + c)2
= a2b2 +2a2bd + 2acbd + d2(a2 + 2ac + c2)
= a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2 [Using (1) and (2)]
= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2
= a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2
[Using (2) and (3) and rearranging terms]
= a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2+ c2 + d2)
= (a2 + b2 + c2) (b2 + c2 + d2)
= L.H.S.
∴ L.H.S. = R.H.S.
open parentheses a squared plus b squared plus c squared close parentheses open parentheses b squared plus c squared plus d squared close parentheses equals open parentheses a b plus b c plus c d close parentheses squared