Prove that square of any term of the sequence 1, 4, 7, 10,... is also the term of the sequence
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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.
The starting term is a1 = 1.
Therefore the n th term , an = a1 + (n-1)d = 1+(n-1)3.
So an = 1+(n-1)3 = 1+3n-3 = 3n-2.
Therefore the nth term an = 3n-2
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