Question

Insert 3 arithmetic means between 6
and 54.​

Answer 1

Solution:

ARITHMETIC SEQUENCE:

Finding the THREE TERMS or 3 arithmetic means between 6 and 54:

Let a2, a3, and a4 be the three terms missing between 6 and 54:

This means that we have 5 terms but we don't know  the 2nd 3rd and 4th terms.  

First thing to do is find first the common difference (d)

That is,

a₅ = a₁ + (n-1)d

54 = 6 + (5-1)d

54 = 6 + 4d

54-6 = 4d

48/4 = 4d/4

d = 12

So we substitute the known values:

a₂ = a₁ + 12

a₂ = 6 + 12  

a₂ = 18

a₃ = a₂ +12

a₃ = 18 +12  

a₃ = 30

a₄ = a₃ +12

a₄ = 30 +12

a₄ = 42

Therefore the sequence are 6, 18, 30 , 42 ,54.

Answer 2

SOLUTION

TO DETERMINE

Insert 3 arithmetic means between 6 and 54.

EVALUATION

We have to insert 3 arithmetic means between 6 and 54.

Let x , y , z are 3 arithmetic means between 6 and 54.

Then 6 , x , y , z , 54 are in Arithmetic Progression

Let common difference = d

So we have

x = 6 + d

y = 6 + 2d

z = 6 + 3d

54 = 6 + 4d

Now 54 = 6 + 4d gives

6 + 4d = 54

⇒ 4d = 48

⇒ d = 12

Thus we have

x = 6 + d = 6 + 12 = 18

y = 6 + 2d = 6 + 24 = 30

z = 6 + 3d = 6 + 36 = 42

FINAL ANSWER

Hence the required 3 arithmetic means between 6 and 54 are 18 , 30 , 42

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