Let T_r be the r th term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, m ≠ n, T_m=1/n and T_n=1/m, - Mathematics MCQ AIEEE 2004 2004 | ExamTest.live

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Let T_{r} be the r th term of an A.P. whose first term is a and common difference is d . If for some positive integers m, n, m \neq = n, T_{m} = \frac{1}{n} and T_{n} = \frac{1}{m}, then a - d equals

Solution & Step-by-step Explanation

In an A.P., the r th term is T_{r} = a + (r - 1) d . Given: T_{m} = a + (m - 1) d = \frac{1}{n} ... (1). T_{n} = a + (n - 1) d = \frac{1}{m} ... (2). Subtracting (2) from (1): (m - n) d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn} d = \frac{1}{mn} . Substitute d into (1): a + (m - 1) \frac{1}{mn} = \frac{1}{n} a = \frac{1}{n} - \frac{m - 1}{mn} = \frac{m - ( m - 1 )}{mn} = \frac{1}{mn} . Since a = \frac{1}{mn} and d = \frac{1}{mn}, a - d = \frac{1}{mn} - \frac{1}{mn} = 0 .

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Let T_{r} be the r th term of an A.P. whose first term is a and common difference is d . If for some positive integers m, n, m \neq = n, T_{m} = \frac{1}{n} and T_{n} = \frac{1}{m}, then a - d equals

0

1

\frac{1}{mn}

\frac{1}{m} + \frac{1}{n}

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