ABCD is a Parallelogram if L and M are the mid-points of BC and DC respectively then $\overrightarrow{\mathrm{AL}}+\overrightarrow{\mathrm{AM}}=\lambda \overrightarrow{\mathrm{AC}}$ then $\lambda$ is

Taking A is the origin, let  the position vectors of B and D be $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{d}}$ respectively Then $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{b}}$and $\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{d}}$

In  $∆$le ABC we have $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}}$
$\begin{array}{l}\Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AD}}\\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{d}}\end{array}$
$\Rightarrow$ position vector of $\mathrm{C}=\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{d}}$
$\therefore$ L and M are mid points of BC and CD Respectively
$\therefore$Position vector of$\mathrm{L}=\overrightarrow{\mathrm{b}}+\frac{(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{d}})}{2}=\overrightarrow{\mathrm{b}}+\frac{1}{2}\overrightarrow{\mathrm{d}}$
Position of vector of $\mathrm{M}=\frac{\overrightarrow{(\mathrm{b}}+\overrightarrow{\mathrm{d}})+\overrightarrow{\mathrm{d}}}{2}=\frac{\overrightarrow{\mathrm{b}}}{2}+\overrightarrow{\mathrm{d}}$
$=\frac{1}{2}\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AD}}$
Now,
$\overrightarrow{\mathrm{AL}}+\overrightarrow{\mathrm{LM}}=(\overrightarrow{\mathrm{b}}+\frac{1}{2}\overrightarrow{\mathrm{d}})+(\frac{1}{2}\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{d}})$
$\begin{array}{l}=\frac{3}{2}\overrightarrow{\mathrm{b}}+\frac{3}{2}\overrightarrow{\mathrm{d}}\\ =\frac{3}{2}(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{d}})\\ =\frac{3}{2}\left(\overrightarrow{\mathrm{AC}}\right)\end{array}$