We want to find a number that belongs to both sequences (it is not necessarily for the same value of n) then we can use n in one term (first one), and m in the other (second one), such that n and m must be integer numbers.

we get:

3n²- 1 = 30–m²

Notice that as n increases, the terms of the first sequence also increase.

And as n increases, the terms of the second sequence decrease.

One way to solve this, is to give different values to m (m = 1, m = 2, etc) and see if we can find an integer value for n.

if m = 1, then:

3n²- 1 = 30–1²

3n²- 1 = 29

3n² = 30

n² = 30/3 = 10

n² = 10

There is no integer n such that n² = 10

now let's try with m = 2, then:

3n²- 1 = 30–2² = 30 - 4

3n²- 1 = 26

3n² = 26 + 1 = 27

n² = 27/3 = 9

n² = 9

n = √9 = 3

So here we have m = 2, and n = 3, both integers as we wanted, so we just found the term that belongs to both sequences.

the number is:

3*(3)² - 1 = 26

30 - 2² = 26

The number that belongs to both sequences is 26.

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2 years ago