Case study:
A school auditorium is to be constructed to accommodate at least 1500 people. The chairs are to be placed in a concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

(i) If the first circular row has 30 seats, how many seats will the 10^th row have?

(ii) For 1500 seats in the auditorium, how many circular rows need to be there?

(iii) If there were 17 rows in the auditorium, how many seats will there be in the middle row?

(i) Given,

The first circular row has 30 seats.

a = 30

Each next row has 10 more seats (d) = 10

So the rows form an A.P.: 30, 40, 50, 60,…

Seats in 10th row :

We know that,

a_n = a + (n - 1)d

⇒ a₁₀ = 30 + (9) × 10

= 30 + 90

= 120.

Hence, there are 120 seats in the 10th row.

(ii) Let n rows be required for 1500 seats.

By formula,

$S_n = \dfrac{n}{2}[2a + (n - 1)d]$

Substituting values, we get :

$\Rightarrow 1500 = \dfrac{n}{2}[2(30) + 10n - 10] \\[1em] \Rightarrow 1500 = \dfrac{n}{2}[60 + 10n - 10] \\[1em] \Rightarrow 1500 = \dfrac{n}{2}[50 + 10n] \\[1em] \Rightarrow 3000 = 10n[5 + n] \\[1em] \Rightarrow \dfrac{3000}{10} = n(n + 5) \\[1em] \Rightarrow 300 = n(n + 5) \\[1em] \Rightarrow 300 = n^2 + 5n \\[1em] \Rightarrow n^2 + 5n - 300 = 0 \\[1em] \Rightarrow n^2 + 20n - 15n - 300 = 0 \\[1em] \Rightarrow n(n + 20) - 15(n + 20) = 0 \\[1em] \Rightarrow (n - 15)(n + 20) = 0 \\[1em] \Rightarrow n - 15 = 0 \text{ or } n + 20 = 0 \\[1em] \Rightarrow n = -20 \text{ or } n = 15.$

Number rows cannot be negative, thus : n = 15

Hence, rows required for 1500 seats = 15.

(iii) If there are 17 rows, seats in the middle row :

Since 17 is odd,

Middle row number = $\dfrac{17 + 1}{2} = \dfrac{18}{2} = 9$.

9th term :

We know that,

a_n = a + (n - 1)d

⇒ a₉ = 30 + (8)10

= 30 + 80

= 110.

Hence, 110 seats will be there in the middle row.