If the sum of $\mathrm{n}$ terms of an $\mathrm{A.P.}$ is $n P+\frac{1}{2} n(n-1) Q,$ where $\mathrm{P}$ and $\mathrm{Q}$ are constants, find the common difference.
If the sum of $\mathrm{n}$ terms of an $\mathrm{A.P.}$ is $n P+\frac{1}{2} n(n-1) Q,$ where $\mathrm{P}$ and $\mathrm{Q}$ are constants, find the common difference.
Let $a_{1}, a_{2}, \ldots a_{n}$ be the given $\mathrm{A.P.}$ Then
${S_n} = {a_1} + {a_2} + {a_3} + \ldots + {a_{n - 1}} + {a_n} = nP + \frac{1}{2}n(n - 1)Q$
Therefore $S_{1}=a_{1}=P, S_{2}=a_{1}+a_{2}=2 P+Q$
So that $a_{2}= S _{2}- S _{1}= P + Q$
Hence, the common difference is given by $d=a_{2}-a_{1}=(P+Q)-P=Q$
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