Solution
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(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
tn=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
⇒ 121−5=3n
⇒ 3n=116
Here, 116 is not divisible by 3
∴ 121 is not a term of the sequence.
(c) Square of nth term =(3n+5)2=9n2+30n+25=(9n2+30n+25)+5
9n2 term and 30n term are divisible by 3 but 20 is not divisible by3. Hence, the square of the nth term will not occur in this sequence.
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