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Arithmetic Progressions
EXERCISE 5.3
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
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?
Step-by-Step Solution
Key Concept: The daily penalties form an Arithmetic Progression (AP) with first term $a = 200$ and common difference $d = 50$. The total penalty for $n$ days is the sum of the first $n$ terms of the AP, given by $S_n = \frac{n}{2}[2a+(n-1)d]$.
1. Identify the AP parameters: - First term $a = 200$ (penalty on the 1st day). - Common difference $d = 250-200 = 50$ (each succeeding day the penalty increases by Rs. 50). - Number of delayed days $n = 30$.
2. Write the formula for the sum of the first $n$ terms of an AP: $$S_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr]$$