Question

An amphitheatre has 50 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. How many seats are there in the 50th row?

Answer 1

Given:

An amphitheater has 50 rows of seats.

To find:

seats are there in the 50th row

Solution:

It is given in the question that there are 20 seats in the first row, 22 seats in the second row, and so on.

so the series for the given condition is given as:

20, 22, 24,..................... Till 50th term

as you can see that the given series is the Arithmetic progression

where the

first term - 20

Common Difference - 22-20 = 2

as according to the question we have to find the seats in the 50th row that means we have to find the value of the 50th term of the given AP

in any of the AP, the nth term is given by:

• aₙ= a+(n-1)d
• a₅₀=20+(50-1)2
• a₅₀ = 20+49×2
• a₅₀ = 20+98
• a₅₀ = 118

There will be 118 seats in the 50th row

Answer 2

$Total \:number \:of \:rows \:in \:an \: amphitheatre = 50$

$Number \:of \: seats \: in \: each \: row$

$20,22,24,26,\ldots up to \:50 \:rows$

$\: respectively \:are \: in \:A.P$

$First \:term (a = a_{1}) = 20$

$Common \: difference (d)$

$= a_{2} - a_{1}$

$= 22 - 20$

$= 2$

/* We know that, */

$\boxed{ \pink{ n^{th} \:term (a_{n}) = a + (n-1)d }}$

$Here, a = 20 , d = 2 \:and \:n = 50$

$Number \:of \:seats \: in \: 50^{th} \:row$

$a_{50} = 20 + (50-1)\times 2$

$= 20 + 49 \times 2$

$= 20 + 98$

$= 118$

Therefore.,

$\red{Number \:of \:seats \: in \: 50^{th} \:row }\green {= 118}$

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