Question

An amphitheatre was planned to be constructed on our school premises such that there wer 38 seats in the front row in the second row, 46 in the third row and so on. The seating was such that there were 4 more seats in each succeeding row. If there were 2016 seats in all calculate the number of rows and the number of seats in the last.

Answer 1

Answer:

24 rows

130 seats in last row

Step-by-step explanation:

first row = 38 seats

2nd row = 42 seats

3rd row = 46 seats

this is in AP

38, 42, 46 ........

a = 38 d = 4

total number of seats = 2016

Sn = n/2 (2a + (n-1)d)

$2016 = \frac{n}{2} (2(38) + (n - 1)4) \\ 2016 = \frac{n}{2} (76 + 4n - 4) \\ 2016 = \frac{n}{2} (72 + 4n) \\ 2016 = \frac{n}{2} \times 2(36 + 2n) \\ 2016 = n(36 + 2n) \\ 2016 = 36n + 2 {n}^{2} \\ 2 {n}^{2} + 36n - 2016 = 0 \\ 2( {n}^{2} + 18n - 1008) = 0 \\ {n}^{2} + 18n - 1008 = 0 \\ {n}^{2} + 42n - 24n - 1008 = 0 \\ n(n + 42) - 24(n + 42) = 0 \\ (n + 42)(n - 24) = 0 \\ n + 42 = 0 \: \: \: \: n - 24 = 0 \\ n = - 42 \: \: \: \: \: \: \: \: \: \: n = 24$

as the number of term cannot be negative

n = 24

therefore the number of rows are 24

number of seats in last row

an = a + (n-1)d

= 38 + (24-1)4

= 38 + (23 × 4)

= 38 + 92

= 130

the number of seats in last row are 130

hope you get your answer

Answer 2

Step-by-step explanation:

Hope you get your answer