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Arithmetic Progressions | Arithmetic Progressions - Exercise 5.3

Question 26

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.

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We can see, that the given penalties are in the form of A.P. having first term as 200 and common difference as 50.

Therefore, a = 200 and d = 50

Penalty that has to be paid if contractor has delayed the work by 30 days = _S_30

By the formula of sum of nth term, we know,

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

Therefore,

S_{30}\\ =\frac{30}{2}[2(200)+(30-1) 50]\\ =15[400+1450]\\ =15(1850)\\ =27750

Therefore, the contractor has to pay Rs 27750 as penalty.

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Our top 5% students will be awarded a special scholarship to Lido.

subject-cta

Question 26

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.

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

We can see, that the given penalties are in the form of A.P. having first term as 200 and common difference as 50.

Therefore, a = 200 and d = 50

Penalty that has to be paid if contractor has delayed the work by 30 days = _S_30

By the formula of sum of nth term, we know,

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

Therefore,

S_{30}\\ =\frac{30}{2}[2(200)+(30-1) 50]\\ =15[400+1450]\\ =15(1850)\\ =27750

Therefore, the contractor has to pay Rs 27750 as penalty.

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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