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a number of linked rings, each 1 cm thick, are hanging on a peg. the top ring has an outside diameter of 20 cm. the outside diameter of each of the outer rings is 1 cm less than that of the ring above it. the bottom ring has an outside diameter of 3 cm. what is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?



Answer :

The distance between the rings linked to each other from the top edge of the top ring to the bottom edge of the bottom ring is 18 cm.

It is given that the rings are linked to each other, hence they do not have any space between them.

Here it is given that from the top ring to the bottom ring, the outside diameter keeps on reducing by 1 cm. This means that this follows an AP i.e Arithmetic Progression from 20cm to 3 cm with a common difference of 1 cm.

Since all the rings are 1 cm thick, th distance from the top edge of the top ring to the bottom edge of the bottom ring will be

no. of rings X 1 cm

Let the no. of rings be n

Since we already know that the rings follow an AP with

the first term, i.e a = 20

the common difference, i.e d = -1

last term, i.e aₙ = 3

The aₙ term of an AP = a + (n - 1)d

Putting the value of aₙ, a, and d we get

3 = 20 + (n - 1)(-1)

or, 3 = 20 - n + 1

or, n = 21 - 3

or, n = 18

Therefore, the distance will be 1cm X 18

= 18cm

To learn more about AP visit

https://brainly.com/question/20343207

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