# The arithmetic mean of three numbers is 23. The second number is six more than the first and the sum of the second and third terms is 54.Calculate the numbers.

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Let the numbers be:

a, b, c

Then:

(a+b+c)/3 = 23

==> a+ b + c = 69 .......(1)

But we know that:

b = a+ 6

==> a = b-6 .........(2)

also:

b+c = 54.........(3)

Now let us substitute (2) and (3) in (1):

a+ b + c = 69

(b-6) + 54 = 69

==> b = 69 - 54 + 6 = 21

==> **b= 21**

==> a = b-6 = 21-6 = 15

==> **a= 15**

==> b+c = 54

==> c = 54- 21 = 54- 21 = 33

==> **c= 33**

We'll note the numbers as a,b,c.

We'll calculate the arithmetic mean:

(a+b+c)/3= 23

We know from enunciation that "The second number is six more than the first" and we'll translate it mathematically as:

b = a + 6 (1)

We also know that "the sum of the second and third terms is 54".

b + c = 54 (2)

Now, we'll substitute the relations (1) and (2) into the arithmetic mean.

(a + 54)/3 = 23

We'll cross multiply:

a+54 = 69

We'll subtract 54 both sides:

a = 69-54

**a = 15**

We'll substitute a in (1):

b = 15 + 6

**b = 21**

We'll substitute b in (2):

b+c = 54

21 + c = 54

c = 54 - 21

**c = 33**

Let the 1st term or number of the AP be x .

The second number of the AP = x+6 by data.

By another condtion, sum of the 1st 2nd and third numbers = 54.

Threfore the 3rd number = 54 -(first +second) = 54 -(x+x+6) = 54-2x-6 = 48-2x.

Since x , x+6 and 48-2x are inAP, they must have the equal common increment between any two successive terms. Or

2nd term -1st term = 3rd term -2nd term

x+6-x = 48-2x -(x+6)

6 = 42 -3x.

3x= 42-6 =36

3x/x = 36/3

x = 12.

So the three numbers in AP are x, x+6 and x+12

and they are 12,18, 24. Obviously they satisfy the condtion. of their sum = 54 .