# 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

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## Related Questions

neela | Student

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 .

giorgiana1976 | Student

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