Answer:
About 630 seventh-grade students would like to snorkel
Step-by-step explanation:
if you do
7/9 x 90 equals 630/810 total students
<span>let x be the interval, then:
186 = 50 + 3 + (3+x) + (3+2x) + (3+3x) + (3+4x) + (3+5x) + (3+6x) + (3+7x)
186 = 74 + 28x
x = 4
Eldest son age = 3+7x = 3+28 = 31.</span>
Let

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

So, the base case is ok. Now, we need to assume
and prove
.
states that

Since we're assuming
, we can substitute the sum of the first n terms with their expression:

Which terminates the proof, since we showed that

as required
Answer:
Step-by-step explanation: