11, 200 logs are stacked in the following manner, 20 logs in the bottom row, 19 in the next row, 18 in
the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in
the top row?
( clear explaination)
Answers
Answer:
Sn=200
AP=20,19,18
a=20
d=19-20= -1
Sn=n/2(2a+(n-1)d)
200=n/2(2(20)+(n-1)-1)
200=n/2(40-n+1)
400=n(41-n)
400=41n-nsquare
41n-nsquare-400=0
nsquare-41n+400=0
nsquare-16n-25n+400=0
n(n-16)-25(n-16)=0
(n-25)(n-16)=0
n=25,16
both will not be the answer
by formula we can find number of rows
an=a+(n-1)d
an=20+(25-1)-1
an=20-24
an= -4 it is impossible
so the number of rows will be 16
please mark me as brainliest
Step-by-step explanation:
Given:-
200 logs are stacked in the following manner, 20 logs in the bottom row, 19 in the next row, 18 in
the row next to it and so on.
To find:-
In how many rows are the 200 logs placed and how many logs are inthe top row?
Solution:-
Total number of logs are stacked = 200
Number of logs in the bottom row=20
Number of logs in the next row = 19
Number of logs in the third row = 18
The number of rows can be written as
20,19,18,....
First term = 20
Common difference = 19-20=-1
d = 18-19=-1
Since the common difference is same throughout the series
They are in the AP
Sum of all logs = 200
=> Sn = 200
We know that
Sum of first n terms in an AP =Sn
Sn = (n/2)[2a+(n-1)d]
On Substituting these values in the above formula
=> (n/2)[2×20+(n-1)(-1)] = 200
=> (n/2)[40+(1-n)] = 200
=> (n/2)[40+1-n]=200
=> (n/2)(41-n)=200
=> (41n-n^2)/2 = 200
=> 41n-n^2 = 2×200
=> 41 n-n^2 = 400
=>n^2-41n+400=0
=>n^2-25n-16n+400=0
=>n(n-25)-16(n-25)=0
=>(n-25)(n-16)=0
=>n-25=0 or n-16=0
=>n=25 or n=16
There can not be 25 rows as we are starting with 20 logs in the first row.
n=16
Number of rows must be 16
Answer:-
Total number of logs in the first row = 16
Used formulae:-
- Sum of first n terms in an AP =Sn
- Sn = (n/2)[2a+(n-1)d]