KnowledgeBoat Logo

Mathematics

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹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?

AP

1 Like

Answer

Since, work has been delayed by 30 days.

Total Penalty (in ₹) : 200 + 250 + 300 + ……… + upto 30 terms

The above sequence is an A.P. with first term (a) = ₹ 200 and common difference = ₹ 50.

By formula,

Sum upto n terms = Sn = n2[2a+(n1)d]\dfrac{n}{2}[2a + (n - 1)d]

Substituting values we get :

S30=302[2×200+(301)×50]=15×[400+29×50]=15×[400+1450]=15×1850=₹ 27750.S_{30} = \dfrac{30}{2}[2 \times 200 + (30 - 1) \times 50] \\[1em] = 15 \times [400 + 29 \times 50] \\[1em] = 15 \times [400 + 1450] \\[1em] = 15 \times 1850 \\[1em] = \text{₹ 27750}.

Hence, total penalty = ₹ 27750.

Answered By

1 Like


Related Questions