Question

if a, b, c, d are in H. P then ab+bc+cd is equal to

Answer 1

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.....