Asked by Jnr John
A display of cans on a grocery shelf consists of 20 cans on the bottom, 18 cans in the next row, and so on in an arithmetic sequence, until the top row has 4 cans. How many cans, in total, are in the display?
Answers
Answered by
Reiny
terms of the sequence:
20 , 18 , 16 ,...4
So we have a = 20, d = -2 , t(n) = 4
t(n) = a + (n-1)d
4 = 20 + (n-1)(-2)
-16 = -2n + 2
2n = 18
n = 9
so we need sum of 9 terms
= (9/2)(first + last)= (9/2)(20 + 4) = 108
or
Sum(n) = (n/2)(2a + (n-1)d)
= (9/2)(40 + 8(-2)) = 108
check:
term9 = a = 8d
= 20 + 8(-2) = 20-16 = 4
20 , 18 , 16 ,...4
So we have a = 20, d = -2 , t(n) = 4
t(n) = a + (n-1)d
4 = 20 + (n-1)(-2)
-16 = -2n + 2
2n = 18
n = 9
so we need sum of 9 terms
= (9/2)(first + last)= (9/2)(20 + 4) = 108
or
Sum(n) = (n/2)(2a + (n-1)d)
= (9/2)(40 + 8(-2)) = 108
check:
term9 = a = 8d
= 20 + 8(-2) = 20-16 = 4
Answered by
Dave
One kind of fruit on the fisrt day of January, two kinds on the second, 3 kinds on the third day, and son how many fruit ate until the twelfh?
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