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. To arrange 10 cans in total, she places 4 cans in the bottom row and 3 cans in the row above it and so on:
i. Alka now arranges a pile having 20 cans in the bottom row, ending with one can in
the top row. The total number of cans in the arrangement are:__________________
ii. There are 820 cans in a pile. How many cans are there in the bottom now?
________________________
iii. Alka has 200 cans. Explain why she will not be able to organize the cans in a
Answers
Answer:
1) 210
2) 40
3) Cannot form
Step-by-step explanation:
We can solve this by number series
1) Last value (l) = 20
First value (a) = 1
Difference between one row and next row (d) = 1
Total no. Of cans = ?
Sum (S) = n/2 [a+l]
n = [(l-a)/d] + 1
= [(20-1)/1] + 1
= 19 + 1
= 20
S = (20/2) (20+1)
= 10 (21)
= 210
So, total no. of in arrangement = 210
2) Total no. cans (S) = 820
a = 1
l = ?
d = 1
n = [(l-a)/d] + 1
= [(l-1)/1] + 1
= l-1+1
n = l
(S) = n/2 [a+l]
(l/2) [1+l) = 820
l (l+1) = 1640
l²+l = 1640
l²+l-1640 = 0
Factorizing it, and we get
(l-40) (l+41)
So, l=40 and l=-41
The negative value not to be considerable. So, l=40
So, the bottom row contains 40 cans.
For check
n =[(a-l)/d] + 1
=[ (40-1)/1] + 1
= 39+1
= 40
(S) = n/2 [a+l] = 820
(40/2) (40+1) = 820
20 (41) = 821
820 = 820
So, answer correct.
3) Total no. cans (S) = 200
Same as previous one. Here also n=l
(S) = n/2 [a+l]
(l/2) (1+l) = 200
l²+l = 400
l²+l-400 = 0
We cannot factorize it. So, with the total of 200 cans, we cannot make a complete pile with given condition.