( 212 + 222 + 232 + . . . + 302 ) = ?
"solution ( 212 + 222 + 232 + . . . + 302 ) = a + ( n - 1 ) d = last number . . . . . so 212 + ( n - 1 ) 10 = 302 . . . . . so n = 10 . so sum = n / 2 ( 2 a + ( n - 1 ) d ) = 10 / 2 ( 2 * 212 + 9 * 10 ) = 2570 . answer c"
divide(add(212, 222), 232)