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Arithmetic Progression 



Problem 16 of 19 


16. 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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Arithmetic Progression 
16. 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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