if the am b/w mth and nth trms of an ap be = to the am b/w pth nd qth terms o an ap ,then show that m+n=p+q

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Solution

$Let a be the first term and d be the common difference . Now, a_{m} = a + (m - 1) d a_{n} = a + (n - 1) d a_{p} = a + (p - 1) d a_{q} = a + (q - 1) d Now, arithmetic mean of a_{m} and a_{n} is given by, A_{1} = \frac{a_{m} + a_{n}}{2} = \frac{a + (m - 1) d + a + (n - 1) d}{2} Now, arithmetic mean of a_{m} and a_{n} is given by, A_{2} = \frac{a_{p} + a_{q}}{2} = \frac{a + (p - 1) d + a + (q - 1) d}{2} Now, A_{1} = A_{2} \Rightarrow \frac{a + (m - 1) d + a + (n - 1) d}{2} = \frac{a + (p - 1) d + a + (q - 1) d}{2} \Rightarrow a + (m - 1) d + a + (n - 1) d = a + (p - 1) d + a + (q - 1) d \Rightarrow (m - 1) d + (n - 1) d = (p - 1) d + (q - 1) d \Rightarrow md - d + nd - d = pd - d + qd - d \Rightarrow (m + n) d = (p + q) d \Rightarrow m + n = p + q$

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