Question

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in 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 be there in the 10^th row? 

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

OR

If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10^th row? 

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

Answer 1

(i) 

Since each row is increasing by 10 seats, so it is an AP with first term a = 30, 

and common difference d = 10. 

So number of seats in 10th row = α₁₀ = a + 9d 

= 30 + 9 x 10 = 120

(ii)

Sn = n² ( 2a + (n-1)d) 

1500 = n/2 ( 2 x 30 + (n-1)10) 

3000 = 50n + 10n² 

n² + 5n - 300 = 0 

n² + 20n -15n – 300 = 0 

(n+20) (n-15) = 0 

Rejecting the negative value, n = 15

OR

No. of seats already put up to the 10th row = S₁₀ 

S₁₀ = 10/2 {2 x 30 + (10-1)10)}

= 5(60 + 90) = 750

So, the number of seats still required to be put are 1500 - 750 = 750

(iii)

If no. of rows = 17 

then the middle row is the 9^th row 

α₈ = a + 8d 

= 30 + 80 

= 110 seats