Answer :

Let a be first term and d be common difference of an AP


Then





Taking LHS



= 6(2a + (n - 1)d


= (12 - 6)(2a + (n - 1)d)


= 3(4 - 2)(2a + (n - 1)d)


= 3[ (4 - 2)(2a + (n - 1)d)]


= 3[ 4(2a + (n - 1)d) - 2(2a + (n - 1)d)]


= 3(S8 - S4) [ By eqn 1 & eqn 2]


= RHS


Hence proved.


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