# A stack of bricks has 61 bricks in the bottom layer, 58 bricks in the second layer, 55 bricks in the third layer,and so on until there are 10 bricks in the last layer. How many bricks are there all together?

Jul 2, 2018

$639$ bricks altogether.

#### Explanation:

The order of bricks is $\left\{61 , 58 , 55 , \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . , 10\right\}$

This is an arithmetic series with first term $a$ as $61$, common difference $d$ as $- 3$ and ${n}^{t h}$ term as $10$

As ${n}^{t h}$ term is $a + \left(n - 1\right) d = 10$, we have

$61 + \left(n - 1\right) \left(- 3\right) = 10$ i.e. $3 \left(n - 1\right) = 61 - 10 = 51$ i.e. $n - 1 = \frac{51}{3} = 17$

and therefore $n = 17 + 1 = 18$

and as sum of the series $\frac{n}{2} \left(a + l\right)$, where $l$ is the last term (observe that it is same as $\frac{n}{2} \left(2 a + \left(n - 1\right) d\right)$)

Number of bricks is $\frac{18}{2} \left(61 + 10\right) = 9 \times 71 = 639$

$639$

#### Explanation:

Number of terms $n$ in the series $61 , 58 , 55 , \setminus \ldots , 10$ with first term $a = 61$ & a common difference $d = - 3$

${a}_{n} = a + \left(n - 1\right) d$

$10 = 61 + \left(n - 1\right) \left(- 3\right)$

$n = 18$

Hence, the total number of bricks in the stack

$= 61 + 58 + 55 + \setminus \ldots + 10$

$= \setminus \frac{n}{2} \left(a + {a}_{n}\right)$

$= \setminus \frac{18}{2} \left(61 + 10\right)$

$= 639$