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In this specific case, the <em>initial term</em> (a) is 5 and the <em>common ratio</em> (r) is -2
Henceforth, after determining what a and r are, we use the formula for the <em>nth term</em>. Which is:
Therefore, the 14th term is :
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Answer:
Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, is equal to
, thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.
Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, . Since a and b were generic elements of H, then H/G is abelian.