if Sn denotes the su of the first n terms of the ap prove that s12=3(s8-s4)
Answers
Answered by
1
hey take this is the correct expanation
Attachments:
Answered by
4
s12=12/2 (2a+(12-1)d)=6 (2a+11d)
s8=8/2 (2a+(8-1)d)=4 (2a+7d)
s4=4/2 (2a+(4-1)d)=2 (2a+3d)
3 (s8-s4)= 3 (4 (2a+7d) - 2 (2a+3d))
=3 (8a+28d - 4 a - 6d)
= 3 ( 4 a + 22d)
= 12a+66d
= 6 (2a+11d)
= s12
so s12= 3 (s8-s4)
s8=8/2 (2a+(8-1)d)=4 (2a+7d)
s4=4/2 (2a+(4-1)d)=2 (2a+3d)
3 (s8-s4)= 3 (4 (2a+7d) - 2 (2a+3d))
=3 (8a+28d - 4 a - 6d)
= 3 ( 4 a + 22d)
= 12a+66d
= 6 (2a+11d)
= s12
so s12= 3 (s8-s4)
Similar questions