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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'

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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^2`


Now, `(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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