Question

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.)

Answer 1

Answer:

The number of seat when n is odd $S_n=\frac{n^2+2n+1}{4}$

The number of seat when n is even $S_n=\frac{n^2+2n}{4}$

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

$S_n=\frac{n}{2}[2a+(n-1)d]$

    $=\frac{n}{2}[a+l]$

a = first term of the series.

d= common difference.

n= number of term

l= last term

$n^{th}$ term of a A.P series is

$T_n=a+(n-1)d$

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 $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

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

$n=1+(t-1)2$

⇒(t-1)2=n-1

$\Rightarrow t-1=\frac{n-1}{2}$

$\Rightarrow t = \frac{n-1}{2}+1$

$\Rightarrow t = \frac{n-1+2}{2}$

$\Rightarrow t = \frac{n+1}{2}$

Last term l= n,, the number of term $=\frac{ n+1}2$, First term = 1

Total number of seat

$S_n=\frac{\frac{n+1}{2}}{2}[1+n}]$

    $=\frac{{n+1}}{4}[1+n}]$

     $=\frac{(1+n)^2}{4}$

    $=\frac{n^2+2n+1}{4}$

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 $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

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

$n=2+(t-1)2$

⇒(t-1)2=n-2

$\Rightarrow t-1=\frac{n-2}{2}$

$\Rightarrow t = \frac{n-2}{2}+1$

$\Rightarrow t = \frac{n-2+2}{2}$

$\Rightarrow t = \frac{n}{2}$

Last term l= n, the number of term $=\frac n2$, First term = 2

Total number of seat

$S_n=\frac{\frac{n}{2}}{2}[2+n}]$

    $=\frac{{n}}{4}[2+n}]$

     $=\frac{n(2+n)}{4}$

    $=\frac{n^2+2n}{4}$  

Answer 2

The arrangement of the seats follow an arithmetic progression.

The number of seats are: $\mathbf{S_n = 2n-n^2}$ and $\mathbf{S_n = 3n -n^2}$

The common difference is given as:

$\mathbf{d = -2}$

The sum of n terms of an arithmetic series is:

$\mathbf{S_n = \frac{n}{2}(2a + (n - 1)d})$

Substitute -2 for d

$\mathbf{S_n = \frac{n}{2}(2a + (n - 1)-2})$

$\mathbf{S_n = \frac{n}{2}(2a -2 (n - 1)})$

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

So, we have:

$\mathbf{S_n = \frac{n}{2}(2(1) -2 (n - 1)})$

$\mathbf{S_n = \frac{n}{2}(2 -2n+2)})$

$\mathbf{S_n = \frac{n}{2}(4 -2n)}$

$\mathbf{S_n = n(2 -n)}$

$\mathbf{S_n = 2n-n^2}$

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

So, we have:

$\mathbf{S_n = \frac{n}{2}(2(2) -2 (n - 1)})$

$\mathbf{S_n = \frac{n}{2}(4 -2 (n - 1)})$

$\mathbf{S_n = \frac{n}{2}(4 -2n + 2})$

$\mathbf{S_n = \frac{n}{2}(6 -2n})$

$\mathbf{S_n = n(3 -n)}$

$\mathbf{S_n = 3n -n^2}$

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