Answer:
Let a be the first term and d be the common difference We know, \[{{S}_{n}}=\frac{n}{2}[2a+(n-1)d]\] Then, \[{{S}_{12}}=\frac{12}{2}[2a+(12-1)]\] \[=6(2a+11d)=12a+66d\] \[{{S}_{8}}=\frac{8}{2}[2a+(8-1)d]\] \[=4(2a+7d)=8a+28d\] and, \[{{S}_{4}}=\frac{4}{2}[2a+(4-1)d]\] \[=2(2a+3d)=4a+6d\] Now, \[3({{S}_{8}}-{{S}_{4}})=3(8a+28d-4a-6d)\] \[=3(4a+22d)\] \[=12a+66d\] \[={{S}_{12}}\] Hence Proved.
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