Question

Clara organizes cans in triangular piles, where each row has one less can than the row below. For example, the pile of 15 cans shown has 5 cans in the bottom row and 4 cans in the row above it.

Answer 1

(a). There will be 80 cans in the bottom row.

(b). n^2 +n - 2S= 0

(c). 2100 cans can't be arranged in a triangular pile.

Step-by-step explanation:

The given question is incomplete ; here is the complete question.

Clara organizes cans in triangular piles, where each row has one less can than the row below. For example, the pile of 15 cans has 5 cans in the bottom row and 4 cans in the row above it.

(a) There are 3240 cans in a pile. How many cans are in the bottom row?

(b) There are S cans and they are organized in a triangular pile with 'n' cans in the bottom row. Show that n^2 +n - 2S= 0

(c) Clara has 2100 cans. Explain why she cant organize them in a triangular pile.

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.

1, 2, 3, 4..........n rows.

(a). Total number of glasses = 3240

Sum of n terms of an arithmetic sequence = $\frac{n}{2}[2a+(n-1)d]$

Where a = first term of the sequence

n = number of rows in a pile

d = common difference

Now, 3240 = $\frac{n}{2}[2(1)+(n-1)(1)]$

6480 = n(n + 1)

n² + n - 6480 = 0

n = $\frac{-1\pm \sqrt{(1)^{2}+4\times 6480}}{2}$

n = $\frac{-1\pm \sqrt{25921}}{2} =\frac{-1\pm 161}{2}$

n = -81, 80

But the number of rows can not be negative therefore, to organize 3240 cans number of rows will be 80.

Since explicit formula of A.P. is $T_{n}=a+(n-1)d$

Number of cans in bottom row,

$T_{(80)}=1+(80-1)(1)=80$ cans

(b). If S can were organized in a triangular pile with n cans in the bottom row, then from the formula

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

$S=\frac{n}{2}[2(1)+(n-1)(1)]$

$S=\frac{n}{2}(n+1)$

n² + n - 2S = 0 will be the equation.

(c). If the number of cans are 2100 then we will find the number of rows to be arranged with these cans

2100 = $\frac{n(n+1)}{2}$

n² + n - 4200 = 0

n = $\frac{-1\pm \sqrt{1^{2}+4\times 4200} }{2}$

n = $\frac{-1\pm \sqrt{8401}}{2}$

n = $\frac{-1\pm 91.66}{2}$

n = 45.32

But the rows can not be in the fraction. Therefore, 2100 cans can't be arranged in a triangular pile.

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Answer 2

Given :  Alka loves to organize cans in triangles piles, where each row has one can less than the  row below and there is only one can in the top row.  

Alka  arranges a pile having 20 cans in the bottom row, ending with one can in  the top row.

To Find : The total number of cans in the arrangement

There are 820 cans in a pile. How many cans are there in the bottom  ?

Alka has 200 cans. Explain why she will not be able to organize the cans in a   triangular pile.

Solution:

Bottom row = 20

Next row = 19

last row = 1

This is an AP

with a = 20

L = 1

d = - 1

n = 20

Sum = (n/2)(a + L )

= (20/2)(1 + 20)

= 210

210 Cans

Let say  n rows

Then bottom row has n  and top row has 1

sum = n(n + 1)/2

n(n + 1)/2 = 820

=> n² + n = 1640

=> n² + 41 n - 40n - 1640 = 0

=> n(n + 41) - 40(n + 41) = 0

=> (n - 40)(n + 41) = 0

=> n = 40

40 cans in bottom row

n(n + 1)/2 =  200

=> n(n + 1) = 400

19(20) = 380

20(21) = 420

so Either 190 or 210 cans be arranged  not 200

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