Sequence is 1,4,7,10,.....
(a) a=1,d=4−1=3
tn=a+(n−1)d
⇒tn=1+(n−1)d
⇒tn=1+3n−3
⇒tn=3n−2 which is algebraic form of the given arithmetic sequence.
(b) tn=3n−2
⇒tn=3n−2
⇒3n=102
⇒n=34
⇒100 is the 34th term of the sequence.
(c) We know, tn=3n−2
Square of the term of this sequence
=(3n−2)2
=9n2−12n+4
=9n2−12n+12−8
=3[3n−2)2−2]−2
Let k=[(3n−2)2−2]
Square of the term of this sequence =3k−2
Since square of the sequence is of the form =3k−2, therefore square of any term of the sequence is a term of this sequence.