# if a, 3, b, 11 is an arthimetical progression, find a and b.if a, 3, b, 11 is an arthimetical progression, find a and b.

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You need to remember that an arithmetic progression relates its terms in the following way such that:

Considering the equation you may evaluate b such that:

Considering the equation , you may substitute 7 for b and then you may finda such that:

**Hence, evaluating a and b under the given conditions yields and .**

a, 3, b, 11 is an arthimatical progression, then:

a1= a

a2= a + r = 3.........(1)

a3= a+ 2r = b ........(2)

a4= a+ 3r = 11.......(3)

Now using the slimination method, subtract (1) from (3):

==> 2r = 8

==> r= 4

Now to find a, from equation (1):

a+ r = 3

==> a + 4 = 3

==> **a= -1**

Now to find b, from (2):

a+ 2r = b

-1 + 2(4) = b

-1 + 8 = b

==> **b= 7**

**Then, **

**-1, 3, 7, 11 are terms in an arthimatical progression where r= 4**

We'll use the theorem of arithmetic mean to determine the terms of the arithmetic progression.

3 = (a+b)/2 => 6 = a+b (1)

b = (3+11)/2

b = 14/2

**b = 7**

We'll substitute b into (1):

6 = a + 7

We'll subtract 7 both sides, to isolate a:

6-7 = a

**a = -1**

We'll verify if the terms a and b are the terms of the a.p. in this way:

3 - a = b - 3 = 11 - b = r

3 - (-1) = 7 - 3 = 11 - 7 = r

4 =4 = 4 = r

**The terms of the arithmetic progression are, whose common difference is r = 4 :**

**-1 , 3 , 7 , 11 , ....**

The terms a , 3, b, 11 are in arithmetic progression.To find a and b.

Solution:

In an arithmetic progression, the consecutive terms has the same common diffrence or

a2-a1 = a3-a2 = a4-a3 , where a1,a2,a3 and a4 are the 1st ,2nd ,3rd and 4th terms of the arithmetic progression in order.

a1 =a, a2=3 , a3 = b and a4 = 11 = d

Therefore .

a2 = a1+d

a4 = a1+3d

a4-a2 = (a+3d-(a1+d) = 2d =11-3 =8 So d = 8/2 = 4,

Therefore a1 = a2- d = 3-4 = -1. Or a = -1.

a3 = b = a2 +4 = 3+4 = 7. Or b = 7.

So a = -1 and b = 7

Let 'an' represent the nth term of the given arithmetic progression. Then:

Also let

d = common difference between two consecutive terms of the series.

Then:

a1 = a

a2 = a1 + d = a + d

a3 =a1 + 2d

a4 = a1 + 3d

Therefore:

a4 - a2 = (a1 + 3d) - (a1 + d) = 2d

However it is given that a2 = 3 and a4 = 11

Substituting these given values of a2 and a4:

a4 - a2 = 11 - 3 = 8 = 2d (as calculated above.

Therefore:

d = 8/2 = 4

Therefore:

a = a2 - d = 3 - 4 = -1

And:

b = a2 + d = 3 + 4 = 7

Answer:

a = -1

b = 7