There are 820 cans in a pile. How many cans are there in the bottom row?
Answers
GIVEN:-
Their are a total 820 cans in a triangular pile. How many cans are there in the bottom row ?
Solution:-
When we count the cans from the top of a triangular pile, we find the number of the cans in each row forming an arithmetic sequence as :- 1, 2, 3, 4n rows.
We have , total cans in the triangular pile are 820 .
in our AP we have ,
- First term = a = 1
- common difference = d = 1
- cans in bottom row = Let n .
- Sn = 820
Putting all values ,
→ Sn = (n/2)[2a + (n - 1)d]
→ 820 = (n/2)[2*1 + (n - 1)1]
→ 1640 = n(2 + n - 1)
→ 1640 = n² + n
→ n² + n - 1640 = 0
→ n² + 41n - 40n - 1640 = 0
→ n(n + 41) - 40(n + 41) = 0
→ (n + 41)(n - 40) = 0
→ n = (-41) and 40
Since value of rows in negative is not Possible..
Therefore, there are 40 cans in the bottom row .
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Question-
There are a total of 820 cans in a triangular pile. How many cans are there in the bottom row?
Answer-
When we count the cans from the top of a triangular pile, we find the number of the cans in each row forming an arithmetic sequence as:-
1, 2, 3, 4 - n rows.
we have, total cans in the triangular pile are 820.
in our AP we have,
- First term = a = 1
- common difference = d = 1
- cans in bottom row = Let n
- Sn = 820
Putting all values,
→ Sn = (n/2)[2a + (n - 1)d]
→ 820 = (n/2)[2*1 + (n - 1)1]
→ 1640 = n(2 + n - 1)
→ 1640 = n² + n
→ n² + n - 1640 = 0
→ n² + 41n - 40n - 1640 = 0
→ n(n + 41) - 40(n + 41) = 0
→ (n + 41)(n - 40) = 0
→ n = (-41) and 40
since the value of rows in negative is not Possible.