Q. 9

# Prove that the sq

a = first term = 4

d = common difference = 7-4 = 3

an = a + (n-1) × d

an = 4 + 3n-3 = 3n + 1

(an)2 = (3n + 1)2 = 9n2 + 6n + 1 = 3n(3n + 2) + 1 = 3{n(3n + 2)} + 1

As, (an)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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