If Sn denots the sum of first n terms of an AP , prove that ,S12=3(S8-S4).
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
Let a be the first term and d be the common difference of the given AP. Then,Sn= {tex}\\frac{n}{2}{/tex}{tex} \\cdot {/tex}[2a+(n-l)d],{tex}\\therefore{/tex}\xa0{tex}3(S_8-S_4) = 3{/tex}[{tex}\\frac{8}{2}{/tex}{tex}(2a+7d)-{/tex}{tex}\\frac{4}{2}{/tex}{tex}(2a+3d)]{/tex}= {tex}3[4(2a+7d)- 2(2a+3d)] = 6(2a+11d){/tex}{tex}= \\frac { 12 } { 2 } \\cdot ( 2 a + 11 d ) = S _ { 12 }{/tex}.Hence, S12= 3(S8-S4).