Seven homogeneous bricks, each having a length L , are arranged as...

Let the mass of each brick be $m$. The origin $O$ is at the left edge of the bottom-most brick.

The $x$-coordinate of the centre of mass of the first (bottom) brick is:
$x_1 = \dfrac{L}{2}$

From the figure, the next three bricks are successively shifted to the right by $\dfrac{L}{10}$. Their centres of mass are:
$x_2 = \dfrac{L}{2} + \dfrac{L}{10}$
$x_3 = \dfrac{L}{2} + \dfrac{2L}{10}$
$x_4 = \dfrac{L}{2} + \dfrac{3L}{10}$

The top three bricks are successively shifted to the left by $\dfrac{L}{10}$. Their centres of mass are:
$x_5 = x_4 - \dfrac{L}{10} = \dfrac{L}{2} + \dfrac{2L}{10}$
$x_6 = x_5 - \dfrac{L}{10} = \dfrac{L}{2} + \dfrac{L}{10}$
$x_7 = x_6 - \dfrac{L}{10} = \dfrac{L}{2}$

The $x$-coordinate of the centre of mass of the entire system of 7 bricks is given by:
$X_{cm} = \dfrac{m(x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7)}{7m}$

Substituting the values:
X_{cm} = \dfrac{1}{7} \left[ 7\left(\dfrac{L}{2}\right) + \dfrac{L}{10}(0 + 1 + 2 + 3 + 2 + 1 + 0) \right]

$X_{cm} =$\dfrac{1}{7} \left[ \dfrac{7L}{2} + \dfrac{9L}{10} \right]$<br><br>$X_{cm} = \dfrac{L}{2} + \dfrac{9L}{70}$<br><br>$X_{cm} = \dfrac{35L + 9L}{70} = \dfrac{44L}{70} = \dfrac{22L}{35}$<br><br>Answer:$\dfrac{22L}{35}