The first row of an auditorium has 34 seats and each next row has 7 more seats. How many seats are there in 21 rows?
Answer and explanation
Correct answer: 2184
This is an arithmetic-progression word problem because the number of seats increases by a constant amount, 7, from one row to the next. Here the first term is a = 34, the common difference is d = 7, and the number of terms is n = 21. The required total is S_n = n/2[2a + (n - 1)d]. Thus S_21 = 21/2[2(34) + 20(7)] = 21/2(68 + 140) = 21/2 × 208 = 2184. Hence option C is correct. Options A, B and D result from incorrect substitution or arithmetic.
Frequently asked questions
What is the correct answer to this question?
2184
Why is this the correct answer?
This is an arithmetic-progression word problem because the number of seats increases by a constant amount, 7, from one row to the next. Here the first term is a = 34, the common difference is d = 7, and the number of terms is n = 21. The required total is S_n = n/2[2a + (n - 1)d]. Thus S_21 = 21/2[2(34) + 20(7)] = 21/2(68 + 140) = 21/2 × 208 = 2184. Hence option C is correct. Options A, B and D result from incorrect substitution or arithmetic.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Word problems based on APs.
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