Question

A gardener plans to construct a trapezoidal shaped structure in his garden. The longer side of trapezoid needs to start with a row of $97$ bricks. Each row must be decreased by $2$ bricks on each end and the construction should stop at ${25}^{th}$ row. How many bricks does he need to buy?

Answer 1

It is given that the longer side of trapezoid needs to start with a row of $97$ bricks and each row must be decreased by $2$ on each end and the construction should stop at $25$th row.

Let us write the number of bricks in each row as an arithmetic sequence as follows:

$97,93,89,......$

Now, we need to find the number of bricks needed to buy and for that we have to write the above sequence as series that is

$97+93+89+......$

The first term of the series is $a_1=97$, thesecond term is $a_2=93$ and so on.

We find the common difference $d$ by subtracting the first term from the second term as shown below:

$d=a_2-a_1=93-97=-4$

We know that the sum of an arithmetic series with first term $a$ and common difference $d$ is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Thus, substitute $a=97,d=-4$ and $n=25$ in $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ as follows:

$S_{25}=\frac{25}{2}\left[\right(2\times 97)+(25-1\left)\right(-4\left)\right]=\frac{25}{2}[194+(24\times -4\left)\right]=\frac{25}{2}[194-96]=\frac{25}{2}\times 98=25\times 49=1225$

Hence, the gardener needs to buy $1225$ bricks.