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 rupees 50 more than for the preceding day. How much money the contractor has to pay, if he has delayed the work by 30 days. SOLVE THIS QUESTION BY THE FORMULA--- Sn= n/2 ( a + l ).

Answer 1

Penalty given for first day= 200

Penalty given for second day = 250

Penalty given for third day = 300

Every day penalty increase = 50

It forms an arithmetic series.

Now

first term a = 200

common difference d = 50

Total number of terms = 30 (because we have to calculate penalty for 30 days)

Now sum 30 terms of AP = (n/2) *{2a + (n-1)*d}

= (30/2)*{2*200 + (30-1)*50}

= 15*(400 + 29*50)

= 15*(400 + 1450)

= 15*1850

= 27750

So total penalty for 30 days = 27750 rupees

IN order to use the formula, you can first find the last term and then substitue.
l = a+(n-1)d = 200 + (30 - 1)50 = 1650
S = n/2 (a+l) = 30 / 2 (200 + 1650 ) = 15 x 1850 = 27750