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

Hint: This question can be done by first determining the first term, common difference of the given arithmetic progression and after that we will apply the $n^{th}$ formula to find the $n^{th}$ term. For this formulas are mentioned below: -
$n^{th}$ term of an A.P (arithmetic progression) is given by ${T_n} = a + (n - 1)d$ where a= first term of the sequence and d=common difference which is given by $d = {T_n} - {T_{n - 1}}$
Sum of n terms is given by ${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$

Complete step-by-step answer:
A) The arithmetic sequence given in this question is 8, 11, 14...Here first term a=8 and common difference ‘d’ is given by $d = {T_n} - {T_{n - 1}}$
$ \Rightarrow d = 11 - 8 = 3$ Here 11 is second term and 8 is first term.
Now $n^{th}$ term of the sequence will be given by ${T_n} = a + (n - 1)d$
$ \Rightarrow {T_n} = 8 + (n - 1)3$ Here a=8 and d=3
Now we will open the brackets and multiply 3 with n-1 terms.
$ \Rightarrow {T_n} = 8 + 3n - 3 = 3n + 5$
Thus the algebraic form of the arithmetic sequence 8, 11, 14... is $3n + 5$
B) Now checking 121 is a term of sequence or not if yes then how. For this we will use $n^{th}$ term formula:-
$ \Rightarrow {T_n} = a + (n - 1)d$
Now we will put 121 in the $n^{th}$ term of the sequence because to prove it is a term of the sequence it must satisfy this $n^{th}$ equation.
$ \Rightarrow 121 = 8 + (n - 1)3$ Here a=8 and d=3
Now we will open the brackets and multiply 3 with n-1 terms.
$ \Rightarrow 121 = 8 + 3n - 3$
$ \Rightarrow 121 = 3n + 5$
Now we will take the variable to one side.
$ \Rightarrow 3n = 121 - 5 = 116$
$ \Rightarrow n = \dfrac{{116}}{3}$
As we can see that 116 is not divisible by 3 therefore 121 is not a term in sequence.
121 is not a term in the sequence.
C) Now we will prove that squares of any term of sequence will not occur in this sequence.
Therefore square of $n^{th}$ term = ${(3n + 5)^2}$
Now we will apply identity ${(a + b)^2} = {a^2} + {b^2} + 2ab$
$ \Rightarrow 9{n^2} + 25 + 30n$
Here we can see that 9 and 30 are divisible by 3 but 25 is not divisible by 3.
Therefore the square of any term of the sequence will not occur in the sequence.

Note: Alternative method to solve third part of the question is that simply take any term for example we will take first term which is 8 whose square term is 64 then if we divide that term by 3 we can clearly see that it is not divisible by 3 so we can say that square term of any sequence will not be in sequence.