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if a, b, c, d are in HP, then prove that ab + bc + cd = 3ad

if a, b, c, d are in HP, then prove that ab + bc + cd = 3ad

Grade:upto college level

6 Answers

san
10 Points
7 years ago
(2ac/(a+c))=b ...1 and (2bd/b+d))=c ...2 [property used=harmonic mean] multiplying one and two 4abcd=bc(a+c)(b+d) 4abcd=ab2c + abcd + b2c2 + c2bd 3abcd=ab2c + b2c2 + c2bd---(3) dividing (3) by bc 3ad=ab+bc +cd can u help me as well my ques is in thermal physics sec title is doubt thanx
Subham Sen
8 Points
7 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...........
Abdul Moiz
13 Points
3 years ago
a,b,c,d are in H.P
so (1/a),(1/b),(1/c),(1/d) are in A.P
d=(1/b)-(1/a)=(1/c)-(1/b)=(1/d)-(1/c)=x..........eq(1)
now (1/b)-(1/a)=(1/c)-(1/b)=(1/d)-(1/c)
=(a-b)/ab=(b-c)/bc=(c-d)/cd=(a-b+b-c+c-d)/(ab+bc+cd) by addendo
=(a-d)/(ab+bc+cd)
taking ad common
(1/b)-(1/a)=ad{(1/d)-(1/a)}/(ab+bc+cd)......... eq(2)
as we know
an=a1+(n-1)d
put n=4
a4=a1+(4-1)d
putting a4=1/d , a1=1/a , d=x
1/d=1/a+3x
put this in eq(2)
(1/b)-(1/a)=ad{(1/a)+3x-(1/a)}/(ab+bc+cd)
x=3adx/(ab+bc+cd)
ab+bc+cd=3ad
Jatin
32 Points
2 years ago
 
a,b,c,d are in H.P
so (1/a),(1/b),(1/c),(1/d) are in A.P
d=(1/b)-(1/a)=(1/c)-(1/b)=(1/d)-(1/c)=x..........eq(1)
now (1/b)-(1/a)=(1/c)-(1/b)=(1/d)-(1/c)
=(a-b)/ab=(b-c)/bc=(c-d)/cd=(a-b+b-c+c-d)/(ab+bc+cd) by addendo
=(a-d)/(ab+bc+cd)
taking ad common
(1/b)-(1/a)=ad{(1/d)-(1/a)}/(ab+bc+cd)......... eq(2)
as we know
an=a1+(n-1)d
put n=4
a4=a1+(4-1)d
putting a4=1/d , a1=1/a , d=x
1/d=1/a+3x
put this in eq(2)
(1/b)-(1/a)=ad{(1/a)+3x-(1/a)}/(ab+bc+cd)
x=3adx/(ab+bc+cd)
ab+bc+cd=3ad
Ankit
13 Points
2 years ago
a b c d are in hp then hp formula.                                   Now using formula b=2ac/a+c.                                 b=2ac/a+c.  equation 1 .                                               C=2bd/b+d equa...2 .                                                     Now eq(1) and eq(2) are multiple.                               Then.  bc= 4abcd/(a+c).(b+d)                                        bc is common so bc cut .                                              4ad=(a+c).(b+d).                                                           4ad=ab+ad+bc+cd.                                                        4ad-ad=ab+bc+cd.                                                           3ad=ab+bc+cd......the proove it.  thanku
ankit singh
askIITians Faculty 614 Points
one year ago
 
a,b,c,d are in H.P
so (1/a),(1/b),(1/c),(1/d) are in A.P
d=(1/b)-(1/a)=(1/c)-(1/b)=(1/d)-(1/c)=x..........eq(1)
now (1/b)-(1/a)=(1/c)-(1/b)=(1/d)-(1/c)
=(a-b)/ab=(b-c)/bc=(c-d)/cd=(a-b+b-c+c-d)/(ab+bc+cd) by addendo
=(a-d)/(ab+bc+cd)
taking ad common
(1/b)-(1/a)=ad{(1/d)-(1/a)}/(ab+bc+cd)......... eq(2)
as we know
an=a1+(n-1)d
put n=4
a4=a1+(4-1)d
putting a4=1/d , a1=1/a , d=x
1/d=1/a+3x
put this in eq(2)
(1/b)-(1/a)=ad{(1/a)+3x-(1/a)}/(ab+bc+cd)
x=3adx/(ab+bc+cd)
ab+bc+cd=3ad

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