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Find the common difference of AP whose \[{{n}^{th}}\] term is xn + y.

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Last updated date: 15th Jul 2024
Total views: 447k
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Answer
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Hint: To solve the question, we have to compare the given series with the general arithmetic progression series to obtain equations of the unknown values. Then solve the obtained equations to calculate the unknown values.

“Complete step-by-step answer:”
We know that the general series of arithmetic progression is given by \[a,\text{ }a+d,\text{ }a+2d,..a+\left( n-1 \right)d\]
Where a, d and n are the first term, the common difference of the series of arithmetic progression and the number of terms in the series of arithmetic progression.
The given \[{{n}^{th}}\] term of arithmetic progression = xn + y
We know from the above series of arithmetic progression that \[{{n}^{th}}\]term of the series is given by the formula a + (n - 1)d
\[\begin{align}
  & xn+\text{ }y=a+\left( n\text{ }-\text{ }1 \right)d \\
 & xn+y-a=\left( n\text{ }-\text{ }1 \right)d \\
\end{align}\]
By rearranging the terms, we get
\[d=\dfrac{xn+y-a}{(n-1)}\] ….. (1)
We know that a is first term of series,
Thus, we get the first term by substituting n = 1 in \[{{n}^{th}}\] term of arithmetic progression = xn + y
The first term fo the given series of AP = x(1) + y = x + y
The value of a = x + y
By substituting the value of a in equation (1), we get
\[d=\dfrac{xn+y-(x+y)}{(n-1)}\]
\[d=\dfrac{xn+y-x-y}{(n-1)}\]
\[d=\dfrac{xn-x}{(n-1)}\]
\[d=\dfrac{x(n-1)}{(n-1)}\]
\[\Rightarrow d=x\]
\[\therefore \] The common difference of the given arithmetic progression = x

Note: The possibility of mistake can be not using the arithmetic progression general series formula to compare and calculate the answer. Using another method of writing series using a pattern method will elongate the process of solving.