Alpha writes the infinite arithmetic sequence \[10, 8, 6, 4, 2, 0 ,\ldots.\] beta writes the infinite geometric sequence \[9, 6, 4, \frac{8}{3}, \frac{16}{9}, \ldots.\] gamma makes a sequence whose $n^{\text{th}}$ term is the product of the $n^{\text{th}}$ term of alpha's sequence and the $n^{\text{th}}$ term of beta's sequence: \[10\cdot 9 \quad,\quad 8\cdot 6\quad ,\quad 6\cdot 4\quad,\quad 4\cdot \frac83\quad,\quad 2\cdot \frac{16}{9}\quad,\quad \ldots.\] what is the sum of gamma's entire sequence?

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

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