Question

Is 210 a term of the A.P $2,6,10,.....?$ If yes, then which term?

Answer 1

Hint: In this question, we are given a sequence and we have been asked to find whether the given number is a term of the sequence or not. It is clearly given that the sequence in the question is an A.P. Now, note the common difference and the first term of the given A.P and use the formula ${a_n} = a + (n - 1)d$ by keeping ${a_n} = 210$ to find whether 210 is a term or not. If it is, then the $'n'$ resulting from the operations will be our answer to the $2^{nd}$ part.

Complete step-by-step answer:
We are given a sequence in this question. Let us note down the first term and the common difference of the sequence.
First term (a) = 2
Common difference (d) = $6 - 2 = 4$
Now we will take ${a_n} = 210$ and put all the terms in the general formula of A.P $ \Rightarrow {a_n} = a + (n - 1)d$
$ \Rightarrow 210 = 2 + (n - 1)4$
Solving for n,
$ \Rightarrow 210 - 2 = (n - 1)4$
Shifting and solving,
$ \Rightarrow \dfrac{{208}}{4} + 1 = n$
$ \Rightarrow 52 + 1 = 53 = n$
Now, since the value of n is a natural number, 210 is a term of the given A.P. And, by our operations we have got $n = 53$.

Hence, 210 is the $53^{rd}$ term.

Note: In this question, if the value of $'n'$ had resulted out to be any whole number, then 184 would have been a term of the given sequence. For example: Let’s find out whether 123 is a term of this sequence or not.
Using the same formula- ${a_n} = a + (n - 1)d$
$ \Rightarrow 123 = 3 + (n - 1)4$
Simplifying to solve for ‘n’,
$ \Rightarrow 123 - 3 = (n - 1)4$
$ \Rightarrow \dfrac{{120}}{4} + 1 = n$
$ \Rightarrow n = 31$
Now, since 31 is a whole number, we can say that 123 is a term of the given sequence. Therefore, $'n'$ needs to be a whole number.