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.
Related Questions
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.
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