Question

for an A.P a=6 and s10 =195show that sn-3:sn-8=n(n-3):(n-5)(n-8).

Answer 1

$first\:\: term ( a ) = 6 \\S_{10}=195$
we know,
$S_n =\frac{n}{2}[2a+(n-1)d]$
where , d is common difference of AP
$So, S_{10}=\frac{10}{2}[2\times6+(10-1)d]\\\\195=5[12+9d]\\39=12+9d\\d=3$
now,
$S_{n-3}= \frac{n-3}{2}[2\times6+(n-3-1)3]\\=\frac{n-3}{2}[12+3n-12]= \frac{3n(n - 3)}{2}$

similarly,
$S_{n-8}=\frac{n-8}{2}[2\times2+(n-8-1)3]\\=\frac{n-8}{2}[12+3n -27]\\=\frac{3(n-8)(n-5)}{2}$
now,
$S_{n-3}:S_{n-8}=n(n-3):(n-5)(n-8)$