Asked by Marcel
You have 798 cans in 19 rows. If the cans are placed so that each row as 3 fewer cans than the other how many cans are in the bottom row?
Answers
Answered by
Reiny
This is an arithmetic series
You know S<sub>19</sub>, n=19, d=-3
You don't know "a".
S<sub>19</sub>= (n/2)[2a + (n-1)d]
798 = 19/2(2a + 18(-3)]
1596 = 38a - 1026
2633 = 38a
a = 69
So depending on whether your bottom row is the
first or the 69th, it would be either 69 or
69 + 18(-3) = 15
check:
69 + 66 + 63 + ... + 15 for 19 terms
= 19(first+last)/2 = (19)(69+15)/2 = 798
You know S<sub>19</sub>, n=19, d=-3
You don't know "a".
S<sub>19</sub>= (n/2)[2a + (n-1)d]
798 = 19/2(2a + 18(-3)]
1596 = 38a - 1026
2633 = 38a
a = 69
So depending on whether your bottom row is the
first or the 69th, it would be either 69 or
69 + 18(-3) = 15
check:
69 + 66 + 63 + ... + 15 for 19 terms
= 19(first+last)/2 = (19)(69+15)/2 = 798
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