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Misc 1 - Show that sum of (m + n)th and (m - n)th terms of AP - Miscellaneous

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Misc 1 Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term. First we calculate (m + n)th , (m – n)th and mth terms of an A.P We know that an = a + (n – 1)d Where an is nth term of AP a be the first term & d be the common difference of the A.P. For (m + n)th term putting n = m + n in (1) am + n = a + (m + n – 1)d For (m – n)th term putting n = m – n in (1) am – n = a + (m – n – 1)d for mth term putting n = m in (1) am = a + (m – 1)d We need to prove Sum of (m + n)th term & (m – n)th terms is twice of mth term i.e. am + n + am – n = 2am Taking L.H.S am + n + am – n = a + (m + n – 1)d +[a + (m – n – 1)d] = a + a + (m + n – 1)d + (m – n – 1)d] = 2a + (m + n – 1 + m – n – 1 )d = 2a + (m + m – 1 – 1 + n – n)d = 2a + (2m – 2 – 0)d = 2a + 2(m – 1)d = 2[a + (m – 1)d] = 2am = R.H.S Hence L.H.S = R.H.S Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.