Answers:
The first three terms are <u>21, 24, 29</u>
<u>5 terms</u> of the sequence less than 50.
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Explanation:
Plug in n = 1 to find that n^2+20 = 1^2+20 = 1+20 = 21. The first term is 21.
Repeat for n = 2 and you should get 2^2+20 = 24 as the second term.
The third term is 29 through similar steps, but this time you use n = 3 of course.
The first three terms are 21, 24, 29
It's effectively the first three perfect squares 1, 4, 9 but we have a tens digit of 2 stuck to the left of each value.
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We want to find when n^2+20 is less than 50. We could keep plugging values of n into that expression and record if the result is less than 50 or not, then tally those occurrences. You should record 5 such occurrences.
Or we could do a bit of algebra like so
n^2+20 < 50
n^2 < 50-20
n^2 < 30
sqrt(n^2) < sqrt(30)
n < sqrt(30)
On a calculator, sqrt(30) is roughly 5.4772 which means n < 5.4772. If n is a a natural number, then the largest it can get is n = 5 to ensure n^2+20 is less than 50.
We can see that,
- n^2+20 = 5^2+20 = 45 when n = 5
- n^2+20 = 6^2+20 = 56 when n = 6
This helps show that n = 5 is the largest n to make n^2+20 < 50 a true statement. This means that there are 5 terms of the sequence less than 50.
