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The mth term of an AP is 1/n and nth term is 1/m then show that sum of mn term is 1/2(mn +1) |
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Answer» Let mth TERM of AP be ‘Am’ and NTH term of AP be ‘An’ Therefore, Am = a+ (m-1)d=1/n ….(i) An = a+(n-1)d=1/m ….(ii) Subtracting equation (ii) from (i) d[(m-1)-(n-1)] = 1/n-1/m, d(m-n) = (m-n)/mn, d = 1/mn ….(iii) Substituting equation (iii) in (i) a+(m-1)/mn = 1/n, a = 1/n[1-(m-1)/m], a = 1/mn ….(IV) Now Amn i.e the mnth term of AP = a+(mn-1)d, Substitute equation (iii) and (iv) in Amn, 1/mn+(mn-1)/mn = 1/mn[1+(mn-1)] = mn/mn = 1, then the mn term = 1 sum of mn term :- Amn = mn/2 ( 2/mn + (mn-1)1/mn) = 1 + (mn)/2 - 1/2 = mn /2 + 1/2 = 1/2 (mn + 1) Step-by-step explanation: |
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