Question

prove that an Arithmetic sequence 5,8,11 contains no perfect square

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Answer 1

Answer:

a = 5

d = 8-5 = 3

an = a + (n-1) × d

an = 5 + (n-1) × 3

∴ an = 3n + 2

Let p be a natural number

p2 = 3n + 2

n =

Now, for all integers from 0 to 9(i.e. p from 0 to 9), n does not come out to be an integer.

Hence, the arithmetic sequence 5, 8, 11, … contains no perfect squares.