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Miscellaneous
Misc 2 Deleted for CBSE Board 2023 Exams
Misc 3 Important Deleted for CBSE Board 2023 Exams
Misc 4 Deleted for CBSE Board 2023 Exams
Misc 5 Deleted for CBSE Board 2023 Exams
Misc 6 Important Deleted for CBSE Board 2023 Exams
Misc 7 Important
Misc 8
Misc 9
Misc 10 Important
Misc 11
Misc 12
Misc 13
Misc 14 Important
Misc 15 Deleted for CBSE Board 2023 Exams
Misc 16 Important Deleted for CBSE Board 2023 Exams
Misc 17
Misc 18
Misc 19 Important
Misc 20
Misc 21 (i) Important Deleted for CBSE Board 2023 Exams
Misc 21 (ii) Deleted for CBSE Board 2023 Exams
Misc 22 Important Deleted for CBSE Board 2023 Exams
Misc 23 Important Deleted for CBSE Board 2023 Exams
Misc 24 Deleted for CBSE Board 2023 Exams
Misc 25 Important Deleted for CBSE Board 2023 Exams
Misc 26 Deleted for CBSE Board 2023 Exams
Misc 27 Deleted for CBSE Board 2023 Exams
Misc 28 Important Deleted for CBSE Board 2023 Exams
Misc 29 Important
Misc 30 Deleted for CBSE Board 2023 Exams
Misc 31 Important
Misc 32 Important Deleted for CBSE Board 2023 Exams
Last updated at March 22, 2023 by Teachoo
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