Question

: A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?​

Answer 1

Step-by-step explanation:

Given

a=200

d=50

n=30

according to Sn=n/2[2a+(n-1)d]

30/2[2(200)+(30-1)50]

15[400+1500-50]

15[1450]

27750 Rs

Answer 2

Hyy dude

Penalty for 1st day = 200

penalty for 2nd day = 250

Penalty for 3rd = 300

Hence , the series is

200 , 250 , 300

State difference is same , It is Ap

Common difference = d = 50

We need to find total penalties work is delayed by 30 days

i.e. we need to find S 30

We know that

Sn =n/2 ( 2a+ ( n - 1 ) d

Here , n = 30 , a =200 , d = 50

Puting these in formula

Sn = n/2 ( 2 a +( n - 1 ) d )

30/2 ( 2 × 200 + (30-1 ) × 50 )

15 ( 400 + 29 × 50 )

15(400 +1450 )

15 × 1850

27750

Here , total penalty if work is delayed by 30 days is rupees 27750

May this helps you

Plz marked in brainlest answer