Question

If abcd are in gp show that (a+b) (b+c)(c+d) are in gp

Answer 1

aka b c d are in gp
b = ak
c = ak^2
d = ak^3
now
a + b = a + ak
b+c = ak + ak^2
c + d = ak^2 +ak^3
now
(b + c)/(a + b) = (ak+ak^2)/(a +ak) =k
(c+d)/(b+c) = (ak^3+ak^2)/(ak+ak^2)=k
therefore
b+c/a+b = c+d/b+c
a+b b+c and c+d are in gp

Answer 2

Step-by-step 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.