# Provide 2 methods of finding variable x is the terms 1,6x+2,13,19,..are terms of an a.p.

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It is given that the AP consists of the terms 1 , 6x + 2 , 13 , 19 ...

As the consecutive terms of an AP have a common difference:

13 - 6x - 2 = 6x + 2 - 1

=> 11 - 6x = 6x +1

=> 12x = 10

=> x = 10 / 12

=> x = 5 / 6

Also, if we take three consecutive terms of an AP, the second term is equal to the average of the 1st and the 3rd terms.

=> (1 + 13)/2 = 6x + 2

=> 14 = 12x + 4

=> 10 = 12x

=> x = 10 / 12

=> x = 5/6

**Therefore the required value of x is 5/6.**

If the given series it's an arithmetical progression, that means that each term, beginning from the second one, is the sum between previous term and a constant amount, called common difference.

We can find the ratio in this way:

19=13+d

19-13=d

d=6

Based on this rule, we can write the term 6x+2 = 1+d

6x+2=1+6

6x = 7 - 2

x = 5/6

Another way of solving the problem would be to consider the term

(6x+2)=(1+13)/2.

6x+2=7

6x=7-2

6x=5

x=5/6

**Both ways, we've get the same value for the variable x, namely x = 5/6.**