Science, asked by Sahifa7098, 1 year ago

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

Answered by aksh173
5
It will help:
(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) 
Answered by Anonymous
0

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

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