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 10th 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 10th row? (iii) If there were 17 rows in the auditorium, how many seats will be there in the middle row

 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 10th row?
(ii) For 1500 seats in the auditorium, how many rows need to be there?

If 1500 seats are to be arranged in the auditorium, how many seats are still left to
be put after 10th row?
(iii) If there were 17 rows in the auditorium, how many seats will be there in the
middle row

 

Arithmetic progression with a=30

d=10

 

number of seats in the 10th row

n=10

tn =a+(n-1) d

t(10) = 30 +(10-1)10 =30+9*10=30+90=120 seats in the 10th row

number of rows required if there is to be a total of 1500 seats

Sn=1500 

n = ?

 

Sn = (n/2){2a+(n-1)d} 

1500=(n/2){2*30+(n-1)*10}

1500=(n/2){60+10n-10}

1500*2=n{50+10n}

3000=50n+10(n^2)

10(n^2)+50n-3000=0

dividing 10

(n^2)+5n-300=0

splitting the middle term and factoring 

(n^2)+20n-15n -300=0

(n+20)(n-15)=0

n= (-15) rejected or n=15 rows

15 rows are required for 1500 seats

if 10 rows are already arranged, 

S(10) = (10/2){2*30+(10-1)*10} = 5*{60+90}=750 seats are already arranged

remaining seats = 1500-750 = 750seats more 

If n=17 rows (odd) 

middle = (n+1)/2 = (17+1)/2 =9th row

t(9) = 30+(9-1)*10 = 30+8*10 = 110seats in the middle 9th row

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