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La suma de los primeros $\\var{n1}$ términos de una progresión aritmética es $\\var {s1}$ y el término de lugar $\\var {n2}$ de la misma serie es $\\var {T}$.

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Recuerde que la fórmula para la suma de los primeros n términos de una progresión aritmética es \$S_n=\\frac{n}{2}(2a+(n-1)d)\$.

La suma de los \$\\var{n1}\$ primeros términos de una progresión aritmética es \$\\var{s1}\$

\$\\frac{\\var{n1}}{2}(2a+\\simplify{{n1}-1}d)=\\var{s1}\$ ecuación (i)

La fórmula para el enésimo término de una progresión aritmética es \$T_n=a+(n-1)d\$.

El término de lugar \$\\var{n2}\$ de la misma serie es \$\\var{T}\$

\$a+\\simplify{{n2}-1}d=\\var{T}\$ ecuación (ii)

Tenemos dos ecuaciones simultáneas. Podemos eliminar el término \$a\$ .

\$\\var{n1}a+\\simplify{({n1}-1)*{n1}/2}d=\\var{s1}\$ ecuación (i)

\$\\var{n1}a+\\simplify{{n1}*({n2}-1)}d=\\simplify{{n1}*{T}}\$ ecuación (ii)*\$\\var{n1}\$

\$\\simplify{({n1}-1)*{n1}/2-{n1}*({n2}-1)}d=\\simplify{{s1}-{n1}*{T}}\$

\$d=\\frac{\\simplify{{s1}-{n1}*{T}}}{\\simplify{({n1}-1)*{n1}/2-{n1}*({n2}-1)}}\$

\$d=\\simplify{({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}\$

Usando este resultado y sustituyendo en la ecuación (ii) podemos encontrar el valor de \$a\$

\$a+\\simplify{({n2}-1)*({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}=\\var{T}\$

\$a=\\var{T}-\\simplify{({n2}-1)*({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}\$

\$a=\\simplify{{{T}-({n2}-1)*({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}}\$

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Calcular el valor de la diferencia común. \$d\$ = [[0]]

Calcular el valor del primer término de la progresión. \$a\$ = [[1]]

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