If the given terms of the series are the consecutive terms of an arithmetical progression, that means that the middle term is the arithmetical mean of the joined terms.

5a+1 = (1+11)/2

We'll remove the brackets:

5a+1 = 12/2

5a+1 = 6

We'll subtract 1 both sides, in order to isolate x:

5a = 6-1

5a = 5

We'll divide by 5 both sides:

**a = 1**

The terms of the arithmetic sequence are:

1, 5*1+1, 11,...

**1 , 6 , 11 ,....**

The common difference of the a.p. is:

6-1 = 11-6 = ... = 5

**d = 5**

The first term of the a.p. is:

** a1 = 1.**

As the terms 1, 5a+1, 11... forms an arithmetic progression we have 1, 5a+1 and 11 as three consecutive terms.

Therefore as with any AP, two times any term is equal to the sum of the terms that lie on either side of it.

We get 1+11= 2* (5a+1)

=> 12=2*(5a+1)

=>6=5a+1

=>5a=5

=>a=1

**Therefore a is equal to 1.**

In an AP the succesive terms have the same common diffrence.

So

2nd term - first term = 3rd term - 2nd term.

5a+1 -1 = 11-(5a+1).

Now we solve for a from this equation :

5a = 11-1-5a.

Add 5a to both sides:

5a+5a = 10

10a = 10

a = 10/10 = 1.

a = 1.