If the m term of an ap is 1/n and the n term is 1/m show that the sum of m and n term is 1/2 ( mn + 1)
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→ aₘ = a + (m - 1)d = 1/n ___(1)
→ aₙ = a + (n - 1)d = 1/m ___(2)
Sₘₙ = (mn)/2 × [2a+(mn-1)d]
Substraction of (2) and (1),
→ nd - md = 1/m - 1/n
→ (n - m)d = (n - m)/mn
→d = 1/mn
→ a + (m - 1)d = 1/n
→ a + (m - 1)1/mn = 1/n
→ a + 1/n - 1/mn = 1/n
→ a = 1/mn = d
→ L.H.S ↓
→ Sₘₙ = (mn)/2 × [2a+(mn-1)d]
→ (mn)/2 × [2/mn+(mn-1)1/mn]
→ (mn)/2 × [2/mn + (mn)/mn - 1/mn]
→ mn/2 × [1/mn + 1]
→ 1/2 + mn/2
→ ½(mn + 1) =R.H.S
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