A theatre has 30 rows of seats arranged in such a way that the number of seats in each row forms an arithmetic progression. The first row contains 18 seats, and each successive row has 3 more seats than the previous row. Due to renovation rules, only those rows whose seat count is divisible by 6 are reserved for premium booking. Find the total number of premium seats in the theatre.
Solution
✅ Correct Option: 1
The number of seats in row $n$ is $a_n=18+3(n-1)=3n+15=3(n+5)$. This is divisible by 6 when $n+5$ is even, so $n$ must be odd. The premium rows are $1,3,5,\ldots,29$, giving 15 rows. Their seat counts form an arithmetic progression from 18 to 102. Total premium seats $=\dfrac{15}{2}(18+102)=900$.
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