S1,S2,S3 are sums of n1,n2,n3 terms of arithmetic sequence. Prove rel S1/n1(n2-n3)+S2/n2(n3-n1)+S3/n3(n1-n2)=0

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You need to remember the formula of sum of n terms of an arithmetic progression such that:

`S_n = (a_1+a_n)*n/2`

The general term of arithmetic progression `a_n`  may be expressed in the following way such that:

`a_n = a_1 + (n-1)d`

`S_n = (2a_1+(n-1)d)*n/2`

You need to write the sum of first `n_1`  terms such that:

`S_1 =(2a_1+(n_1 - 1)d)*n_1/2`

Hence, evaluating `S_1/n_1`  yields:

`S_1/n_1 = ((2a_1+(n_1 - 1)d)*n_1/2)/n_1`

`S_1/n_1 = (2a_1+(n_1 - 1)d)/2`

Reasoning by analogy yields:

`S_2/n_2 = (2a_1+(n_2 - 1)d)/2`

`S_3/n_3 = (2a_1+(n_3 - 1)d)/2`

`(S_1/n_1)(n_2-n_3) + (S_2/n_2)(n_3-n_1) + (S_3/n_3)(n_1-n_2) = (2a_1+(n_1 - 1)d)/2(n_2-n_3) + (2a_1+(n_2 - 1)d)/2(n_3-n_1) + (2a_1+(n_3 - 1)d)/2(n_1-n_2)`

`(S_1/n_1)(n_2-n_3) + (S_2/n_2)(n_3-n_1) + (S_3/n_3)(n_1-n_2) = a_1(n_2-n_3 + n_3-n_1 + n_1-n_2) + (d/2)((n_1 - 1)(n_2-n_3) + (n_2 - 1)(n_3-n_1) + (n_3 - 1)(n_1-n_2))`

Reducing equal terms in the first group yields:

`(S_1/n_1)(n_2-n_3) + (S_2/n_2)(n_3-n_1) + (S_3/n_3)(n_1-n_2) = a_1*0 + (d/2)(n_1n_2 - n_1n_3 - n_2 + n_3 + n_2n_3 - n_2n_1 - n_3 + n_1 + n_1n_3 - n_2n_3 - n_1 + n_2)`

Reducing equal terms in the second group yields:

`(S_1/n_1)(n_2-n_3) + (S_2/n_2)(n_3-n_1) + (S_3/n_3)(n_1-n_2)`  `= a_1*0 + (d/2)*0 `

`(S_1/n_1)(n_2-n_3) + (S_2/n_2)(n_3-n_1) + (S_3/n_3)(n_1-n_2) = 0`

Hence, evaluating the given expression, using the formula of sum of n terms of arithmetic progression, yields `(S_1/n_1)(n_2-n_3) + (S_2/n_2)(n_3-n_1) + (S_3/n_3)(n_1-n_2) = 0.`

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