find the seating capacity of a movie house with 40 rows of seats if there are 25 seats in the first row, 28 seats in the second row, 31 in third row and so on.
Answers
Answer:
3340 seats
Step-by-step explanation:
Given data:
In a movie house, there are 40 rows in total with 25 seats in the first row, 28 seats in the second row and 31 seats in the third row and so on.
To find:
The capacity of the movie house
Step-by-step explanation:
Clearly the given problem is based on Arithmetic Progression since 28 - 25 = 3 = 31 - 28. We form the A.P. as follows,
25, 28, 31, 34, 37, ... ... (upto 40 terms)
Then first term = 25 and common difference = 3
Now the sum of the 40 terms of the A.P.
= 40/2 * [2 * 25 + (40 - 1) * 3]
= 20 * [50 + 39 * 3]
= 20 * [50 + 117]
= 20 * 167
= 3340
Final answer:
The capacity of the movie house is 3340.
NOTE:
If a be the first term of an A.P. and d be the common difference, then the sum of n numbers of terms is given by
Sn = n/2 * [2a + (n - 1) d]
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