For an A.P. a=6 and S10=195
Show that
Sn-3:Sn-8=n(n-3):(n-5)(n-8).
Answers
Answered by
4
Sn=n/2(2a+(n-1)d)
S10=10/2(2a+(10-1)d)
195=5(2(6)+9d) (substitute a=6)
39=(12+9d)
9d=39-12=27
d=27/9=3
Sn-3:Sn-8 =(n-3/2(2(6)+(n-3-1)3):(n-8/2(2(6)+(n-8-1)3)
=(n-3/2(12+3n-12)):(n-8/2(12+3n-27))
=(n-3/2(3n)):(n-8/2(3n-15))
=(n-3)(n):(n-8)(n-5)
Hence proved
I hope its useful
S10=10/2(2a+(10-1)d)
195=5(2(6)+9d) (substitute a=6)
39=(12+9d)
9d=39-12=27
d=27/9=3
Sn-3:Sn-8 =(n-3/2(2(6)+(n-3-1)3):(n-8/2(2(6)+(n-8-1)3)
=(n-3/2(12+3n-12)):(n-8/2(12+3n-27))
=(n-3/2(3n)):(n-8/2(3n-15))
=(n-3)(n):(n-8)(n-5)
Hence proved
I hope its useful
Similar questions