Question

Is $68$ a term of the A.P. $7,10,13,...$ ?

Answer 1

Hint: We need to check whether $68$ is a term of the A.P. $7,10,13,...$ . First, let $68$ be the \[{{n}^{th}}\] term of the given A.P. We will use the equation ${{a}_{n}}=a+(n-1)d$ to find the value of $n$ . Find $a$ and $d$ from the given A.P. and substitute in the equation. If $n$ is a whole number, then $68$ is a term of the A.P. $7,10,13,...$ .

Complete step-by-step answer:
We need to find whether $68$ is a term of the A.P. $7,10,13,...$ .
For this, let us assume $68$ to be the \[{{n}^{th}}\] term of the given A.P., i.e ${{a}_{n}}=68$ .
We know that ${{a}_{n}}=a+(n-1)d...(i)$ , where $a$ is the first term of the A.P., $d$ is the common difference and $n$ is the number of terms.
Now we have to find \[a\] and $d$ .
$a$ is the first term of the A.P. $=7$ .
$d$ is the common difference \[={{a}_{2}}-{{a}_{1}}\] .
$\Rightarrow d=10-7=3$
Now let’s substitute these values in equation $(i)$ to find the value of $n$ so that we can find which term of the A.P. is $68$.
Therefore, $7+(n-1)3=68$
Now, rearrange the terms to obtain the value of $n$ .
$\Rightarrow 7+3n-3=68$
Now, collect variable terms to one side and constant terms to the other
$\Rightarrow 3n=68-4$
$\Rightarrow 3n=64$
$\Rightarrow n=\dfrac{64}{3}$
Since $n$ is a term, it cannot be a fraction.
Therefore, $68$ is not a term of the A.P. $7,10,13,...$ .

Note: In questions of these types, take the given term as the \[{{n}^{th}}\] term and evaluate the value of $n$. $n$ must be a whole number. It cannot be a fraction, integer or any other type. Let us see an alternative solution to the given question.
From the A.P. $7,10,13,...$ , we can see that each term of the A.P. has a difference of $3$ .
Consider the first term, i.e. $7$ . This can be written as $(3\times 2)+1$ .
Similarly, $10$ can be written as $(3\times 3)+1$ .
So $13$ can be written as $(3\times 4)+1$ .
Now, the nearest term $68$ to that is a multiple of $3$ is $66$ and $69$.
So the two adjacent terms nearer to $68$ will be $(3\times 22)+1=66+1=67$ and $(3\times 23)+1=69+1=70$
So it is impossible to get $68$ as a term.