The lengths of medians of the given triangle are used to form a second triangle ,then the medians of that one are used to form a third triangle, and so on. Find the sum of the areas of all the triangles in that sequence. Area of the original triangle is A.
Please help me with any suggestion or help to solve this problem.
Let us give a simple geometric argument. Consider first the triangle $\Delta ABC$, draw its mass center $G$, the intersection of the medians, and then consider the parallelogram formed by the two segments $AG$ and $CG$, and the two parallel ones, $CM$ and $AM$ in the picture.
Now the medians divide the area of the triangle $\Delta ABC$ in six (area) equivalent pieces. (For instance, the median from $A$ divides the triangle in two equivalent pieces. Then $BG$ divides the one half in the proportion $2/3$ to $1/3$, which is the same proportion for $G$ dividing the median. The $2/3$ part of the half is now divided in two equivalent parts by the median from $G$ in $\Delta ABG$. There are many similar arguments.)
Let us denote by $A'$ the area of each of the six equivalent parts, $A=6A'$. Now the triangle $\Delta AGM$ is congruent to a triangle with sides $AG$, $BG$, $CG$, so similar to a triangle $\Delta A_1B_1C_1$, say, with sides congruent to the medians of the initial triangle, the
similitude factor being $2/3$. So $$ \operatorname{Area}(A_1B_1C_1) = \left(\frac 32\right)^2 \operatorname{Area}(AGM) = \left(\frac 32\right)^2 \cdot 2A' = \left(\frac 32\right)^2 \cdot 2\cdot \frac A6 =\frac 34A\ . $$ The same factor $q=\frac 34$ appears when passing to the "next triangle", and so on, so the problem wants us to compute the geometric sum $$ A\left(1+q+q^2+q^3+\dots\right)=A\cdot\frac 1{1-q}=4A\ . $$