Answer :

Let a be the first term and d be the common difference of A.P.

We know that


…………….(1)


…………….(2)


…………….(3)


To prove: S12 = 3(S8-S4)


Solving RHS,


RHS = 3(S8-S4)


= 3[(8a + 28d)-(4a + 6d)] (From eq (2) and (3))


= 3[4a + 22d]


= 12a + 66d


From eq(1),


RHS = 12a + 66d = S12 = LHS


Hence proved.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses