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| Question 864515:  Dominoes are stacked on top of each other such that the first row contains 80, the 2nd row 78, etc so that each row contains 2 fewer dominoes than the previous row. If the number of dominoes in the pile totals 710, how many rows are there in the stack?
 Found 2 solutions by  ewatrrr, richwmiller:
 Answer by ewatrrr(24785)
      (Show Source): Answer by richwmiller(17219)
      (Show Source): 
You can put this solution on YOUR website! d=-2 not +2 710 = ½(160+ (n-1)(-2))n
 710 = ½(160-2n+2))n
 n^2-81n+710 = 0
 n = 10
 S = ½(2a + (n-1)d)n
 S = ½(2*80 + (9)-2)10
 S = ½(160 -18)10
 S = 5*(142)
 s=710
 
 80,78,76,74,72,70,68,66,64,62
 80+78+76+74+72+70+68+66+64+62 = 710
 an = a + (n - 1)d
 a10 = 80 + (9)(-2)=62
 
 n = 71
 S = ½(160-2*(70))71
 S=1/2(20)*71
 s=10*71
 A71 = 80 + (70)-2
 A71=80-140
 a71=-60
 n=71 is an extraneous root
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