SOLUTION: given sequence 2;5;8 2.1.1 if the pattern continues , then write down the next two terms 2.2.2 prove that none of the terms of this sequence are perfect squares

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Question 1177105: given sequence 2;5;8
2.1.1 if the pattern continues
, then write down the next two terms
2.2.2 prove that none of the terms of this sequence are perfect squares

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
given sequence 2;5;8
2.1.1 if the pattern continues, then write down the next two terms
as you can see, common difference is 3, so next two terms are 11 and 14

2.2.2 prove that none of the terms of this sequence are perfect squares
Let each term of the given arithmetic progression be defined as
T%5Bn%5D=T%5B1%5D%2B%28n-1%29d+, where T%5B1%5D=2 and d=T%5B2%5D-T%5B1%5D=5-2=3+
T%5Bn%5D=2%2B3%28n-1%29+
T%5Bn%5D=3n-1+
Hence, each term of the AP is of the form 3n-1,nN.
We know that any integer is of the form 3a+, +3a%2B1 or +3a-1 for +aZ.
Taking x=3a, we have x%5E2=9a%5E2=3%283a%5E2%29=3m , with m=3a%5E2.
Taking +x=3a%2B1, we have +x%5E2=9a%5E2%2B6a%2B1=3%283a%5E2%2B2a%29%2B1=3m%2B1, where m=%283a%5E2%2B2a%29+
Taking x=3a%2B2, we have x%5E2=9a%5E2%2B12a%2B4=3%283a%5E2%2B4a%2B1%29%2B1=3m%2B1, with m=3a%5E2%2B4a%2B1+
Hence, we can see that a perfect square is always of the form 3m+ or 3m%2B1 , where m is an integer.
Hence, we can infer that a perfect square is not of the form 3m-1.
Also, we have each term of the given AP to be of the form 3m-1.
Hence, no+term of the given AP is a perfect+square.