Question

Prove that the squares of all terms of the sequence 4, 7, 10 includes in it.

Answer 1

Step-by-step explanation:

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.

Answer 2

Step-by-step explanation:

Given sequence is 1, 4, 7,10. a1 = 1, a2 = 4, a3 = 7 , a4 = 10. We see that a2 -a1 = 4-1 =3. , a3-a2 = 7-4 = 3, a4 - a3 = 10-7 = 3. Therefore the successive terms have the same difference or common difference d = 3.

-Abhay sir (M.sc., B.ed.)

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