\frac{ \left( 5+5+ \left( n-1 \right) d \right) n }{ 2 } =390
Réitigh do d.
d=-\frac{10\left(n-78\right)}{n\left(n-1\right)}
n\neq 1\text{ and }n\neq 0
Réitigh do n. (complex solution)
\left\{\begin{matrix}n=\frac{\sqrt{d^{2}+3100d+100}+d-10}{2d}\text{; }n=\frac{-\sqrt{d^{2}+3100d+100}+d-10}{2d}\text{, }&d\neq 0\\n=78\text{, }&d=0\end{matrix}\right.
Réitigh do n.
\left\{\begin{matrix}n=\frac{\sqrt{d^{2}+3100d+100}+d-10}{2d}\text{; }n=\frac{-\sqrt{d^{2}+3100d+100}+d-10}{2d}\text{, }&d\leq -20\sqrt{6006}-1550\text{ or }\left(d\neq 0\text{ and }d\geq 20\sqrt{6006}-1550\right)\\n=78\text{, }&d=0\end{matrix}\right.
Tráth na gCeist
\frac{ \left( 5+5+ \left( n-1 \right) d \right) n }{ 2 } =390
Roinn
Cóipeáladh go dtí an ghearrthaisce
\left(5+5+\left(n-1\right)d\right)n=390\times 2
Iolraigh an dá thaobh faoi 2.
\left(10+\left(n-1\right)d\right)n=390\times 2
Suimigh 5 agus 5 chun 10 a fháil.
\left(10+nd-d\right)n=390\times 2
Úsáid an t-airí dáileach chun n-1 a mhéadú faoi d.
10n+dn^{2}-dn=390\times 2
Úsáid an t-airí dáileach chun 10+nd-d a mhéadú faoi n.
10n+dn^{2}-dn=780
Méadaigh 390 agus 2 chun 780 a fháil.
dn^{2}-dn=780-10n
Bain 10n ón dá thaobh.
\left(n^{2}-n\right)d=780-10n
Comhcheangail na téarmaí ar fad ina bhfuil d.
\frac{\left(n^{2}-n\right)d}{n^{2}-n}=\frac{780-10n}{n^{2}-n}
Roinn an dá thaobh faoi n^{2}-n.
d=\frac{780-10n}{n^{2}-n}
Má roinntear é faoi n^{2}-n cuirtear an iolrúchán faoi n^{2}-n ar ceal.
d=\frac{10\left(78-n\right)}{n\left(n-1\right)}
Roinn 780-10n faoi n^{2}-n.
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