
Is 302 a term of the A.P. 3, 8, 13, …..?
Answer
557.7k+ views
Hint: Observe the given A.P. and find its common difference by subtracting the first term from the second term. Now, to check if 302 is a term of the given A.P or not, assume 302 as the \[{{n}^{th}}\] term of the A.P. denoted by \[{{T}_{n}}\]. Apply the formula for \[{{n}^{th}}\] term of an A.P. given as: - \[{{T}_{n}}=a+\left( n-1 \right)d\], where ‘a’ is the first term and ‘d’ is the common difference. Substitute all the values and find the value of ‘n’. If the value of ‘n’ is a whole number then 302 will be a term of the S.P. and if ‘n’ is a fraction then 302 will not be a term of the A.P.
Complete step-by-step solution
Here, we have been provided with an A.P. having terms 3, 8, 13, …… We have to check whether 302 will be a term of the A.P. or not.
Now, let us denote the first term of the A.P. with ‘a’, common difference with ‘d’ and \[{{n}^{th}}\] term with \[{{T}_{n}}\]. On observing the above A.P we can say that: -
\[\begin{align}
& \Rightarrow a=3 \\
& \Rightarrow d={{T}_{2}}-{{T}_{1}}=8-3=5 \\
\end{align}\]
So, let us assume that 302 is the \[{{n}^{th}}\] term of the A.P. denoted with \[{{T}_{n}}\]. So, we have,
\[\Rightarrow {{T}_{n}}=302\]
Now, we know that the \[{{n}^{th}}\] term of an A.P. is given by the relation: - \[{{T}_{n}}=a+\left( n-1 \right)d\]. So, we have,
\[\Rightarrow a+\left( n-1 \right)d=302\] - (1)
Now, if 302 is a term of the A.P. then the value of n must turn out to be a whole number. So, let us check. Therefore, substituting the values of ‘a’ and ‘b’ in equation (1), we get,
\[\begin{align}
& \Rightarrow 3+\left( n-1 \right)5=302 \\
& \Rightarrow \left( n-1 \right)5=299 \\
& \Rightarrow \left( n-1 \right)=\dfrac{299}{5} \\
& \Rightarrow n=\dfrac{299}{5}+1 \\
& \Rightarrow n=\dfrac{304}{5} \\
\end{align}\]
Here, ‘n’ is a fraction which cannot be possible because the number of terms cannot be a fraction. So, it is thereby concluded that 302 is not a term of the given A.P.
Note: One may note that we cannot write all the terms of the given A.P. to check whether 302 is a term of the given A.P. or not because it will take a long time to check-in that manner. So, the formula and process that we have applied in the above solution are preferred. Now, while calculating the common difference ‘d’ we have to consider the two adjacent terms of the A.P. like: - \[{{T}_{2}}-{{T}_{1}},{{T}_{3}}-{{T}_{2}}\], etc. Never consider the difference of terms like \[{{T}_{3}}-{{T}_{1}}\] or \[{{T}_{4}}-{{T}_{2}}\] etc.
Complete step-by-step solution
Here, we have been provided with an A.P. having terms 3, 8, 13, …… We have to check whether 302 will be a term of the A.P. or not.
Now, let us denote the first term of the A.P. with ‘a’, common difference with ‘d’ and \[{{n}^{th}}\] term with \[{{T}_{n}}\]. On observing the above A.P we can say that: -
\[\begin{align}
& \Rightarrow a=3 \\
& \Rightarrow d={{T}_{2}}-{{T}_{1}}=8-3=5 \\
\end{align}\]
So, let us assume that 302 is the \[{{n}^{th}}\] term of the A.P. denoted with \[{{T}_{n}}\]. So, we have,
\[\Rightarrow {{T}_{n}}=302\]
Now, we know that the \[{{n}^{th}}\] term of an A.P. is given by the relation: - \[{{T}_{n}}=a+\left( n-1 \right)d\]. So, we have,
\[\Rightarrow a+\left( n-1 \right)d=302\] - (1)
Now, if 302 is a term of the A.P. then the value of n must turn out to be a whole number. So, let us check. Therefore, substituting the values of ‘a’ and ‘b’ in equation (1), we get,
\[\begin{align}
& \Rightarrow 3+\left( n-1 \right)5=302 \\
& \Rightarrow \left( n-1 \right)5=299 \\
& \Rightarrow \left( n-1 \right)=\dfrac{299}{5} \\
& \Rightarrow n=\dfrac{299}{5}+1 \\
& \Rightarrow n=\dfrac{304}{5} \\
\end{align}\]
Here, ‘n’ is a fraction which cannot be possible because the number of terms cannot be a fraction. So, it is thereby concluded that 302 is not a term of the given A.P.
Note: One may note that we cannot write all the terms of the given A.P. to check whether 302 is a term of the given A.P. or not because it will take a long time to check-in that manner. So, the formula and process that we have applied in the above solution are preferred. Now, while calculating the common difference ‘d’ we have to consider the two adjacent terms of the A.P. like: - \[{{T}_{2}}-{{T}_{1}},{{T}_{3}}-{{T}_{2}}\], etc. Never consider the difference of terms like \[{{T}_{3}}-{{T}_{1}}\] or \[{{T}_{4}}-{{T}_{2}}\] etc.
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