Question

Prove that the squares of all the terms of these 1, 4, 7....... belong the same sequence?

Answer 1

Given AP is $1,\ 4,\ 7,\,\dots$

First let us find $n^{th}$ term of this sequence.

Common difference, $d=4-1=3.$

First term, $a=1.$

Then $n^{th}$ term is,

$\longrightarrow a_n=a+(n-1)d$

$\longrightarrow a_n=1+(n-1)\,3$

$\longrightarrow a_n=1+3n-3$

$\longrightarrow a_n=3n-2$

Let us take square of $n^{th}$ term.

$\longrightarrow (a_n)^2=(3n-2)^2$

$\longrightarrow (a_n)^2=9n^2-12n+4$

$\longrightarrow (a_n)^2=9n^2-12n+6-2$

$\longrightarrow (a_n)^2=3(3n^2-4n+2)-2$

Taking $3n^2-4n+2=n',$

$\longrightarrow (a_n)^2=3n'-2$

This means the square of $n^{th}$ term of our AP is the $(n')^{th}$ term of our AP.

Thus the squares also belong to the same AP.

Hence Proved!