If {S_n} = nP + frac{1}{2}n(n - 1)Q,where {S_ | vedclass

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(D) The sum of the first $n$ terms of an $A.P.$ is given by the formula: ${S_n} = \frac{n}{2} \{ 2a + (n - 1)d \}$,where $a$ is the first term and $d$

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$Q$

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(D) The sum of the first $n$ terms of an $A.P.$ is given by the formula: ${S_n} = \frac{n}{2} \{ 2a + (n - 1)d \}$,where $a$ is the first term and $d$ is the common difference.Given the expression: ${S_n} = nP + \frac{1}{2}n(n - 1)Q$.We can rewrite this as: ${S_n} = \frac{n}{2} \{ 2P + (n - 1)Q \}$.Comparing this with the standard formula,we identify that the common difference $d = Q$.Alternatively,we can find the common difference using the terms:${S_1} = P + 0 = P$${S_2} = 2P + \frac{1}{2}(2)(1)Q = 2P + Q$${T_1} = {S_1} = P$${T_2} = {S_2} - {S_1} = (2P + Q) - P = P + Q$Common difference $d = {T_2} - {T_1} = (P + Q) - P = Q$.

If ${S_n} = nP + \frac{1}{2}n(n - 1)Q$,where ${S_n}$ denotes the sum of the first $n$ terms of an $A.P.$,then the common difference is

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