Subham Sen
Last Activity: 11 Years ago
(1/b)-(1/a)=(1/c)-(1/b)=(1/d)-(1/c)=x(say).........................................(1)
=>(a-b)/ab=(b-c)/bc=(c-d)/cd=(a-b+b-c+c-d)/(ab+bc+cd)....................[By addendo]
so,
(1/b)-(1/a)=(a-d)/(ab+bc+cd)
=>(1/b)-(1/a)=ad{(1/d)-(1/a)}/(ab+bc+cd) [taking `ad` common]
......................................(2)
now
t(n)=a+(n-1)d
or, 1/d = (1/a)+(4-1)x [since , a,b,c,d are in h.p.]
[also from eqn (1), d=(1/b)-(1/a)=x]
so, 1/d = (1/a)-3x
Now substituting this value in eqn (2) we get ,
(1/b)-(1/a)=ad{(1/a)+3x-(1/a)}/(ab+bc+cd)
or, x = 3adx/(ab+bc+cd)
thus
ab+bc+cd=3ad
thanx
n hey , the sum was cool...........