# Determine the real number x, if x+3,x+9 and 3x+11 are 3 consecutive terms of an arithmetical progression.

*print*Print*list*Cite

x+3, x+9, 3x+11

are 3 consecutive terms in arithmetical progression:

==> x+9 = x+3 +r ...(1)

ans 3x+11 = x+9 +r ...(2)

From (1), reduce x :

==> 9= 3+r

==> r= 6

Now, from (2):

3x+11= x+9 +r = x+9+6 = x+15

3x+11 = x+15

2x= 4 ==> x=2

To check:

x+3, x+9, 3x+11

5, 11, 17 are 3 consecutive terms in arithmetical progression with ratio of 6

In an arithmetic progression the difference between any two consecutive terms is same. Thus given n1, n2 and n3 as three consecutive terms in an arithmetic progression:

n2 - n1 = n3 - n2 = c

Thus:

n1 = n1

n2 = n1 + c

n3 = n2 + c = n1 + c = c = n +2c

In the given problem:

n1 = x + 3

n2 = x + 9

therefore :

c = n2 - n1 = x + 9 - x - 3 = 6

And

n3 = n1 + 2c = x + 3 + 2*6 = x + 15

It is given:

n3 = 3x + 11

Therefore:

x + 15 = 3x + 11

3x - x = 15 - 11

2x = 4

x = 2

Answer:

x = 2

Since x+3, x+9 and 3x+11 are the 3 consecutive terms, the common difference is the same from first term and 2nd term and the second term and third term.

So Common difference = (x+9)-(x+3) = (3x+11)-(x+9)

6 = 2x+2. Or

6-2 =2x. Or

4 = 2x. Or x =4/2 = 2

So x= 2 and the three consecutive numbers are x+3 = 5, x+9 = 11 and 2x+11 = 3*2+11= 17

If x+3,x+9 and 3x+11 are consecutive terms of an arithmetical progression, that means that:

x+9 = x+3+r, where r is the ratio

We'll move all terms to one side:

x+9-x-3-r=0

6-r=0

We'll add r both sides:

**r=6**

We also know that:

3x+11 = x+9+r , where r = 6

We'll substitute r by 6 and we'll get:

3x+11 = x+9+6

3x+11 = x+15

We'll move all terms to one side:

3x+11-x-15=0

2x-4=0

We'll add 4 both sides:

2x=4

We'll divide by 2:

**x = 2**