Ramkali saved 5 Rs in the first week of a year and then increased her weekly savings by 1.75 Rs. If in the nth week, her weekly savings become 20.75 Rs, find n.
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Answers
Answer:
10 th week
Step-by-step explanation:
savings in the 1st week = 5 Rs
from the second week onward, the weekly savings incresed by 1.75 Rs
total savings in the 2nd week = 5 Rs + 1.75 = 6.75 Rs
total savings in the 3rd week = 6.75 + 1.75 = 8.50 Rs
So it goes on adding...
first factor = a=5
difference = d =1.75
=20.75
=a+(n−1)d
20.75 = 5+ (n−1) 1.75
15.75 = (n−1) 1.75
(n−1)=
=
= 63/7
= 9
∴n−1 = 9
∴n=10
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Answer :
Given :-
Ramkali saved Rs.5 in the first week of a year and then increased her weekly savings by Rs. 1.75 . In the nth week, her weekly savings become Rs. 20.75
To Find:-
Find the value of n?
Solution :-
The saving of Ramkali in the first week = Rs. 5
The amount of saving is increasing in every week = Rs. 1.75
The saving of second week = 5+1.75
= Rs. 6.75
If we continue like this we get the sequence 5 , 6.75 , 8.50 ,...
First term = (a) = 5
Common difference = (d)
= 6.75-5 = 1.75
= 8.50-6.75 = 1.75
Since the common difference is same throughout the sequence
They are in the AP
Now
Her savings in the nth week = Rs. 20.75
Let the nth term = 20.75
We know that
The general term of an AP = an
an = a+(n-1)d
=> an = 20.75
=> a+(n-1)d = 20.75
On Substituting these values in the above formula then
=> 5+(n-1)×1.75 = 20.75
=> 5+1.75n -1.75 = 20.75
=>1.75 n + 3.25 = 20.75
=> 1.75n = 20.75-3.25
=> 1.75n = 17.50
=> n = 17.50/1.75
=> n = [(1750)/100]/[175/100)]
=> n = 1750/175
=> n = 10
Therefore, n = 10
In 10 th week Ramkali's saving will be Rs. 20.75
Answer:-
The value of n for the given problem is 10
Used formulae:-
- The general term of an AP=a+(n-1)d