Question

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

Answer 1

Answer:

Since a, b, c, d are in HP, 1/a, 1/b, 1/c and 1/d are in AP.

D = 1/b−1/a = 1/c−1/b = 1/d−1/c

where D is the common difference of the AP: 1/a, 1/b, 1/c.

Now,

ab = a−b/D

bc = b−c/D

cd = c−d/D

Adding all we get,

ab+bc+cd = (a−b+b−c+c−d)/D = (a-d)/D --------(1)

Now,

D = 1/3(1/d-1/a)

D = 1/3(a-d)/ad

(a-d)/D = 3ad ------------(2)

putting (2) in (1) we get,

ab+bc+cd = 3ad (Ans)

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Answer 2

Answer:

K=3

Step-by-step explanation:

If a, b, c, d are in H.P

1a,1b,1c,1d1a,1b,1c,1d  are in A.P

Let Common difference be ‘m’

1b−1a=m⇒a−bm=ab1b−1a=m⇒a−bm=ab

1c−1b=m⇒b−cm=bc1c−1b=m⇒b−cm=bc

1d−1c=m⇒c−dm=cd1d−1c=m⇒c−dm=cd

ab+bc+cd=a−bm+b−cm+c−dmab+bc+cd=a−bm+b−cm+c−dm

⇒ab+bc+cd=a−dm⇒ab+bc+cd=a−dm

1d−1a=3m⇒a−dm=3ad1d−1a=3m⇒a−dm=3ad

∴ab+bc+cd=3ad