Question

$a,b,c,d,e$ are in $A.P$. Prove the following results
$a-4b+6c-4d+e=0$

Answer 1

$\begin{array}{c}a-4b+6c-4d+e\\ 2b=a+c\\ 2c=b+d\\ 2d=ec\\ a-4b+6c-4d+e\\ a+e-2\left(2b+2d\right)+6c\\ a+c-4\left(2c\right)+6c\\ a+e-2c\\ a+e-\left(d+b\right)\\ letdiffrent=t\\ d+t=e\\ a+t=b\\ a+d+t-d-a-t\\ =0\end{array}$