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stira [4]
4 years ago
15

If 1/a, 1/b, 1/c are in Arithmetic Progress, show that 2ac/a+c.​

Mathematics
1 answer:
Iteru [2.4K]4 years ago
6 0
  1. 1/b-1/a = 1/c-1/b [ since they are in arithmetic progress]
  2. 1/b+1/b = 1/c+1/a
  3. 2/b = (a+c)/ac
  4. 2ac = b(a+c)
  5. 2ac/(a+c) =b

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mart [117]

Answer:

Heyy mate... here's ur ans ;)

they asked us to apply pythagarus theorem becoz that theorem proves

that it is rectangle so lets get started

( p - perpendicular / b - base / h - hypotenuse )

a) =>

B = 8.1 m

P = 6.3 m

H = 10.15m

Now applying thm => H^2=p^2+b^2

(8.1)^2 + (6.3)^2 = (10.15)^2

65.61 + 39.69 = 103.02

so now LHS = 105.3 and RHS = 103.02 hence

LHS not equal to RHS

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b) =>

B = 14.4

P = 5.2

H = 15.31

Now applying thm => H^2=p^2+b^2

(5.2)^2 + (14.4)^2 = (15.31)^2

27.04 + 207.36 = 234.39

LHS = 234.4 and RHS = 234.39

just taking approximation rhs is 234.4

hence

LHS = RHS

so it is a rectangle

please mark it brainliest <3

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