Q. 9

# Prove that the squares of all the terms of the arithmetic sequence 4, 7, 10, … belong to the sequence.

Answer :

a = first term = 4

d = common difference = 7-4 = 3

a_{n} = a + (n-1) × d

a_{n} = 4 + 3n-3 = 3n + 1

(a_{n})^{2} = (3n + 1)^{2} = 9n^{2} + 6n + 1 = 3n(3n + 2) + 1 = 3{n(3n + 2)} + 1

As, (a_{n})^{2} is also of the form 3n + 1, the squares of all the terms of the arithmetic sequence 4, 7, 10, … belong to the sequence.

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In the table below, some arithmetic sequences are given with two numbers against each. Check whether each belongs to the sequence or not.

Kerala Board Mathematics Part-1

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ii)

Kerala Board Mathematics Part-1

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Kerala Board Mathematics Part-1