39a=3
b=4
c=6
d=12
ab+bc+cd=12+24+72=108
3ad=3x3x12=108
so ab+bc+cd=3ad
3 Answers
1@L , u had been told to prove that statement, u had simply put the values......others try to prove it
1As a,b,c,d are in HP , so 1/a , 1/b , 1/c , and 1/d are in AP.
Let r be its common diff. so that 1/a +r = 1/b , 1/a + 2r =1 / c , 1/a + 3r =1/d )
so 1/b - 1/a =r
so (a-b)/ab = r
so ab = (a-b)/r
similarly bc = (b-c)/r
and cd = (c-d)/r
adding , ab+bc+cd = 1/r ( a-d) ----- (1)
now 1/d - 1 / a = 3r
so (a - d) = 3r x ad ------ (2)
so from (1) and (2) , ab+bc+cd = 3ad