IIT JEE , AIEEE Forum - Mathematics , Algebra

krish1092

If $\frac{bc}{ad}=\frac{b+c}{a+d}=\frac{3(b-c)}{a-d}$ , then a,b,c,d are in

a)A.P

b)G.P

c)H.P

d)none

krish1092

some one please reply

reddevil_2009

i am trying yaar.................

abhiforiit

r they in H.P ???????

krish1092

Please post the method

sahilgupta_iit

I got it in the first try!

see, bc/ad = b+c/a+d

so, 1/b + 1/c = 1/a + 1/d......(1).....which satisfies one of the condtitions for a,b,c,d to be in HP

now bc/ad = 3(b-c)/a-d

.... 3(1/c - 1/b) = 1/d - 1/a.........(2)

add (1) and (2)

4/c - 2/b = 2/d

2/c = 1/b + 1/d......(3)

Now subtract (2) from (1)

4/b - 2/c = 2/a..........(4)

2/b = 1/a +1/c........(5)

so, equations (1), (3), (4), (5) show that a,b,c,d are in HP

Done!

konichiwa2x

The question isnt that complicated. When faced with an HP question, most students start with b = 2ac/a+c or 1/b-1/a=1/c-1/b. The important thing to keep in mind is the relation between distant terms (for eg. 1/d - 1/a = 3(1/c - 1/b) = 2(1/c-1/a) and so on...

If you bear this in mind always, one look at the first and 3rd term will tell u that it is an HP.

abhiforiit

ans) bc/ad = b+c/ a+d = > abc + bcd = abd + adc ----------(1)
then when bc/ad = 3 ( b-c) / a-d => abc -bcd = 3abd - 3adc -----(2)
(1) + (2) = abc = 2abd - adc = solvin v get bc+dc=2bd
which implies b,c,d r in h.p

then agian solve for a,b ,c using any of the 2 equalities u get ab+bc=2ac ....... therefore in H.P

rate if useful !!!!!!!1

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If , then a,b,c,d are in

a)A.P

b)G.P

c)H.P

d)none

If $\frac{bc}{ad}=\frac{b+c}{a+d}=\frac{3(b-c)}{a-d}$ , then a,b,c,d are in