Question

(a) Write algebraic form of the arithmetic sequence $8,11,14,.....$
(b) Is $121$ a term of this sequence? Why?
(c) Prove that the square of any term of this sequence will not occur in this sequence.

Answer 1

(a) Arithmetic sequence is $8,11,14,....$
First term, $a=8$ and common difference, $d=11-8=3$
Thus, nth term of the sequence is
$t_n=a+(n-1)d=8+(n-1)3=8+3n-3=3n+5$
Thus, the algebraic form of the given arithmetic sequence is $3n+5$
(b) Consider the number $121$
Therefore, $121=3n+5$
$\Rightarrow$ $121-5=3n$
$\Rightarrow$ $3n=116$
Here, $116$ is not divisible by $3$
$\therefore$ $121$ is not a term of the sequence.
(c) Square of nth term $={(3n+5)}^2=9n^2+30n+25=(9n^2+30n+25)+5$
$9n^2$ term and $30n$ term are divisible by $3$ but $20$ is not divisible by$3$. Hence, the square of the nth term will not occur in this sequence.