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 If Sn is sum of n even terms of arithmetic progression series and Sn' is sum of n odd terms of arithmetic progression series then prove that Sn = (1 + 1/n) Sn' [ Calculator, Method and examples ]Solution:Your problem -> If Sn is sum of n even terms of arithmetic progression series and Sn' is sum of n odd terms of arithmetic progression series then prove that Sn = (1 + 1/n) Sn'Here S_n = 2 + 4 + 6 + ... + 2n:. S_n = n/2 [ 2a + (n - 1) d ]= n/2 [ 2(2) + (n - 1) * 2 ] (because a = 2 and d = 2)= n/2 [ 4 + 2n - 2 ]= n/2 [ 2n + 2 ]= n ( n + 1 ) Now, S_(n') = 1 + 3 + 5 + ... + (2n - 1):. S_(n') = n/2 [ 2a + (n - 1) d ]= n/2 [ 2(1) + (n - 1) * 2 ] (because a = 1 and d = 2)= n/2 [ 2 + 2n - 2 ]= n/2 [ 2n ]= n^2Now, (S_n)/(S_n') = (n (n + 1))/(n^2) :. (S_n)/(S_n') = ((n + 1))/n:. S_n = (1 + 1/n) × S_(n') (Proved)

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