
An A.P. consists of 31 terms. If 16th term is n, find the sum of all terms of the A.P.
Answer
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Hint: We will assume the first term and common difference of the A.P. as variable. Then we will use the formula of \[{{n}^{th}}\] term of A.P. and the sum of first n terms of A.P. which is as follows:
\[{{n}^{th}}\] term of A.P. \[={{T}_{n}}=a+(n-1)d\]
Sum of first n terms \[=\dfrac{n}{2}\left[ 2a+(n-1)d \right]\], where a and d are the first term and the common difference respectively of the A.P.
Complete step-by-step answer:
We have been given that an A.P. consists of 31 terms,
\[\Rightarrow \] n = 31
Also, we have been given that 16th term of the A.P. is n.
Let us assume the first term and common difference to be a and d respectively of the A.P.
We know that the \[{{n}^{th}}\] term is given by,
\[\begin{align}
& \Rightarrow {{T}_{n}}=a+(n-1)d \\
& \Rightarrow {{T}_{16}}=a+(16-1)d \\
& \Rightarrow {{T}_{16}}=a+15d \\
& \Rightarrow n=a+15d....(1) \\
\end{align}\]
We also know that the sum of the first ‘n’ terms of an A.P. is equal to ‘n’ divided by 2 times the sum of twice the first term and product of the common difference and n minus 1.
\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]\]
So the sum of the first 31 terms is \[{{S}_{31}}\].
\[\Rightarrow {{S}_{31}}=\dfrac{31}{2}\left[ 2a+(31-1)d \right]=\dfrac{31}{2}\left[ 2a+30d \right]\]
Taking 2 as common, we get as follows:
\[\Rightarrow {{S}_{31}}=\dfrac{31}{2}\times 2\left[ a+15d \right]\]
Now, by substituting the value of \[\left[ a+15d \right]\] from the equation (1), we get as follows:
\[\begin{align}
& \Rightarrow {{S}_{31}}=\dfrac{31}{2}\times 2\times n \\
& \Rightarrow {{S}_{31}}=31n \\
\end{align}\]
Hence the sum of all the terms of the A.P.is equal to 31n.
Note: Take care of the sign mistakes while calculation of \[{{T}_{n}}\] and \[{{S}_{n}}\] of an A.P. Also, remember the formulas of \[{{T}_{n}}\] and \[{{S}_{n}}\] of an A.P. as it will help in the problems related with it. If we make a mistake in writing the formula of either of these, we will get the wrong answer.
\[{{n}^{th}}\] term of A.P. \[={{T}_{n}}=a+(n-1)d\]
Sum of first n terms \[=\dfrac{n}{2}\left[ 2a+(n-1)d \right]\], where a and d are the first term and the common difference respectively of the A.P.
Complete step-by-step answer:
We have been given that an A.P. consists of 31 terms,
\[\Rightarrow \] n = 31
Also, we have been given that 16th term of the A.P. is n.
Let us assume the first term and common difference to be a and d respectively of the A.P.
We know that the \[{{n}^{th}}\] term is given by,
\[\begin{align}
& \Rightarrow {{T}_{n}}=a+(n-1)d \\
& \Rightarrow {{T}_{16}}=a+(16-1)d \\
& \Rightarrow {{T}_{16}}=a+15d \\
& \Rightarrow n=a+15d....(1) \\
\end{align}\]
We also know that the sum of the first ‘n’ terms of an A.P. is equal to ‘n’ divided by 2 times the sum of twice the first term and product of the common difference and n minus 1.
\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]\]
So the sum of the first 31 terms is \[{{S}_{31}}\].
\[\Rightarrow {{S}_{31}}=\dfrac{31}{2}\left[ 2a+(31-1)d \right]=\dfrac{31}{2}\left[ 2a+30d \right]\]
Taking 2 as common, we get as follows:
\[\Rightarrow {{S}_{31}}=\dfrac{31}{2}\times 2\left[ a+15d \right]\]
Now, by substituting the value of \[\left[ a+15d \right]\] from the equation (1), we get as follows:
\[\begin{align}
& \Rightarrow {{S}_{31}}=\dfrac{31}{2}\times 2\times n \\
& \Rightarrow {{S}_{31}}=31n \\
\end{align}\]
Hence the sum of all the terms of the A.P.is equal to 31n.
Note: Take care of the sign mistakes while calculation of \[{{T}_{n}}\] and \[{{S}_{n}}\] of an A.P. Also, remember the formulas of \[{{T}_{n}}\] and \[{{S}_{n}}\] of an A.P. as it will help in the problems related with it. If we make a mistake in writing the formula of either of these, we will get the wrong answer.
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