For A.P if a=6,S4=105 then prove that Sn:Sn-3=(n+3):(n-3).

Question

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Answer 1

The nth term for the series is a+(n-1)d where d is the common difference.

S₁=a=6.

So S₄=105=6+3d, 3d=105-6=99, d=33. The series is 6, 39, 72, 105, 138, 171, 204, ...

S_n=6+33(n-1)=6+33n-33=33n-27.

S_n-3=33(n-3)-27=33n-99-27=33n-126. S_n-3=S_n-99.

S_n/S_n-3=(33n-27)/(33n-126)=(11n-9)/(11n-42).

This is not (n+3)/(n-3) generally.

Let n=4, then S₄/S₁=105/6=35/2. But (n+3)/(n-3)=7.

If (11n-9)/(11n-42)=(n+3)/(n-3), then:

(11n-9)(n-3)=(11n-42)(n+3),

11n²-42n+27=11n²-9n-126,

27+126=42n-9n,

153=33n, n=153/33=51/11. So since n must be an integer, S_n:S_n-3≠(n+3):(n-3) for any integer n>3.