consider the arithmetic sequence 9,16,23,...
a) what is the common difference?
b) What is it's algebraic form?
c) prove that the squares of all term of this sequence belong to it;
Answers
Answer:
(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
⇒ 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
=9n
2
+30n+25=(9n
2
+30n+25)+5
9n
2
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.
Answer:
In this arithmetic sequence, the common difference
d
is 7. You can find this by subtracting a term from the consecutive term: 16 - 9 = 7, 23 - 16 = 7.
In order to find the next four terms, keep adding 7 to the previous term.
23 + 7 = 30
30 + 7 = 37
37 + 7 = 44
44 + 7 = 51
The next four terms are 30, 37, 44, and 51