Question

For an A.P. a=6 and S10=195
Show that
Sn-3:Sn-8=n(n-3):(n-5)(n-8).

Answer 1

So hence it prove hope u get it please share it and comments.

Answer 2

Answer:

It is proved that Sn-3:Sn-8=n(n-3):(n-5)(n-8).

Step-by-step explanation:

It is given that a=6 and S₁₀ = 195.

The sum of n terms is

$S_n=\frac{n}{2}[2a+(n-1)d]$

$S_{10}=\frac{10}{2}[2a+(10-1)d]$

$195=5(2(6)+9d)$

$39=12+9d$

$27=9d$

$d=3$

We have to prove Sn-3:Sn-8=n(n-3):(n-5)(n-8).

$S_{n-3}=\frac{n-3}{2}[2(6)+(n-3-1)3]\Rightarrow \frac{n-3}{2}(12+3n-12)=\frac{3n(n-3)}{2}$

$S_{n-8}=\frac{n-8}{2}[2(6)+(n-8-1)3]\Rightarrow \frac{n-8}{2}(12+3n--27)=\frac{3(n-5)(n-8)}{2}$

$\frac{S_{n-3}}{S_{n-8}}=\frac{\frac{3n(n-3)}{2}}{\frac{3(n-5)(n-8)}{2}}=\frac{n(n-3)}{(n-5)(n-8)}$

Hence proved that Sn-3:Sn-8=n(n-3):(n-5)(n-8).