Theater Seating

$The\quad number\quad of\quad seats\quad is\quad in\quad the\quad sequence\\ 15,\quad 18,\quad 21,\quad ...\\ \\ In\quad the\quad sequence\quad the\quad 1st\quad term\quad or\quad { a }_{ 1 }\quad is\quad equal\quad to\quad 15.\\ The\quad common\quad difference\quad or\quad d\quad is\quad equal\quad to\quad 3.\\ \\ The\quad number\quad of\quad seats\quad must\quad be\quad 870,\quad which\quad means\\ the\quad sum\quad of\quad the\quad sequence\quad up\quad to\quad the\quad nth\quad term\\ must\quad be\quad 870.\\ \\ We\quad will\quad use\quad the\quad formula\\ \qquad { S }_{ n }=\frac { n[2{ a }_{ 1 }+(n-1)d] }{ 2 } \\ Where\quad { S }_{ n }\quad is\quad the\quad sum\quad of\quad the\quad seats\\ and\quad n\quad is\quad the\quad number\quad of\quad rows.\\ \\ { a }_{ 1 }=15\\ d=3\\ \\ { S }_{ n }=\frac { n(30+3n-3) }{ 2 } =870\\ \\ \qquad \qquad 3{ n }^{ 2 }+27n-1740=0\\ \qquad \qquad { n }^{ 2 }+9n-580=0\\ \qquad \qquad (n-20)(n+29)=0\\ \qquad \qquad n=20\quad (n>0)\\ \\ \boxed { \therefore \quad The\quad architect\quad must\quad use\quad 20\quad rows\quad in\quad his\quad design. } \\ \\$

n = number of rows

5+6+7+8+...+(n+4) = 290

Solve: n=20