A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the preceding row. Find a formula, S(n), for the number of seats used in the layout. (Hint: The number of seats in the layout depends on whether n is odd or even.)

The arrangement of the seats follow an arithmetic progression.

The number of seats are: and

The common difference is given as:

The sum of n terms of an arithmetic series is:

Substitute -2 for d

When n is odd, the last seat would be 1.

So, we have:

When n is even, the last seat would be 2.

So, we have:

Read more about the sum of arithmetic terms at:

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The number of seat when n is odd

The number of seat when n is even

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

a = first term of the series.

d= common difference.

n= number of term

l= last term

term of a A.P series is

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let of the series is n.

Here , n=t, a=1 and d=2

⇒(t-1)2=n-1

Last term l= n,, the number of term , First term = 1

Total number of seat

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let of the series is n.

Here , n=t, a=2 and d=2

⇒(t-1)2=n-2

Last term l= n, the number of term , First term = 2

Total number of seat

idk

step-by-step explanation:

because 1.2 isnt half of 6 duh