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Question
Write the nth term of the \[A . P . \frac{1}{m}, \frac{1 + m}{m}, \frac{1 + 2m}{m}, . . . .\]
Sum
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Solution
Given:
\[A . P . \frac{1}{m}, \frac{1 + m}{m}, \frac{1 + 2m}{m}, . . . .\]
We know that the nth term of an AP is given by \[a_n = a + \left( n - 1 \right)d\]
In the given AP
In the given AP
\[a = \frac{1}{m}\]
\[d = \frac{1 + m}{m} - \frac{1}{m} = \frac{1 + m - 1}{m} = 1\]
Thus, the nth term of the given AP is
\[d = \frac{1 + m}{m} - \frac{1}{m} = \frac{1 + m - 1}{m} = 1\]
\[a_n = \frac{1}{m} + \left( n - 1 \right)1 = \frac{1 + \left( n - 1 \right)m}{m}\]
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