#### If $S\textsubscript{n}$ denotes the sum of first n terms of an AP, prove that $\\ S\textsubscript{12} = 3(S\textsubscript{8} - S\textsubscript{4})\\$

$\\\text{To prove}:- S\textsubscript{12} = 3(S\textsubscript{8} - S\textsubscript{4})\\ If \ S\textsubscript{n} \text{is the sum of n terms then first term is a and common difference is d} \\ S_{n}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \ldots \ldots .. (1)\\ \text{Put }n = 12 in equation (1) \\ S\textsubscript{12} = 6[2a + 11d]\\ S\textsubscript{12} = 12a + 66d \ldots \ldots \ldots \ldots . (1)\\ S\textsubscript{8} = 4[2a + 7d]\\ S\textsubscript{8} = 8a + 28d \ldots \ldots \ldots \ldots \ldots .(2)\\ S\textsubscript{4} = 2[2a + 3d]\\ S\textsubscript{4} = 4a + 6d \ldots \ldots \ldots \ldots \ldots ..(3)\\ RHS \\3(S\textsubscript{8} - S\textsubscript{4}) = \\ =3(8a + 28d - 4a - 6d) \text{By Equation (2) and (3)}\\ =3(4a + 22d)\\ = 12a + 66d By Equation (1)\\ = S\textsubscript{12}\\ \text{Hence } \textsubscript{12} = 3(S\textsubscript{8} - S\textsubscript{4})$