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✌️11th and 15th....✌️
Penalty for 2nd day = Rs.250
Penalty for 3rd day = Rs.300; and so on.
Total penalty = Rs.27750
This penalty is an A.P. with a = 200,
d (common difference) = 50
Let the work be completed after n days
Sn = 27750
Since the number of days cannot be negative,
n = 30 days
Sn = 4n -n^2
Just Think The Given Equation Of The Sum Of " n " Terms Is correct for all the terms in the AP , So If you put n=1 You get Sum of 1 terms which is actually the value of first term a1 .
It goes like this ,
Sn = 4n - n^2
= 4(1) - (1)^2
= 4 - 1 = 3
so The First Term Is 3.
S2 = 4(2) - (2)^2 = 4
Sum of first 2 Terms means a1 + a2 ,
Now you know a1 and S2 , so S2 - a1 = a2 = 4-3= 1
Now the common difference ,
a2- a1 = d = 1-3 = -2.
Hence The Basic Method Of Doing That was this .
Solving for a10 , an , a3 we get ,
a3 = -1 ;
a10 = -15;
an = 5 - 2n ;
Answers
Answer:
Step-by-step explanation:
15)Penalty for 1st day = Rs.200
Penalty for 2nd day = Rs.250
Penalty for 3rd day = Rs.300; and so on.
Total penalty = Rs.27750
This penalty is an A.P. with a = 200,
d (common difference) = 50
Let the work be completed after n days
Sn = 27750
Since the number of days cannot be negative,
n = 30 days
11) We have been given ,
Sn = 4n -n^2
Just Think The Given Equation Of The Sum Of " n " Terms Is correct for all the terms in the AP , So If you put n=1 You get Sum of 1 terms which is actually the value of first term a1 .
It goes like this ,
Sn = 4n - n^2
= 4(1) - (1)^2
= 4 - 1 = 3
so The First Term Is 3.
Sum Of 2 Terms Is ...
S2 = 4(2) - (2)^2 = 4
Sum of first 2 Terms means a1 + a2 ,
Now you know a1 and S2 , so S2 - a1 = a2 = 4-3= 1
Now the common difference ,
a2- a1 = d = 1-3 = -2.
Hence The Basic Method Of Doing That was this .
Solving for a10 , an , a3 we get ,
a3 = -1 ;
a10 = -15;
an = 5 - 2n ;