Basically the later number = consecutive earlier number in the sequence times 2/3 then subtracts by 4
Sn
Sn+1 = 2/3 Sn-4
Sn+2=2/3Sn+1−4 = 2/3(2/3Sn -4)-4 = 4/9Sn - 20/3
===> Sn = (Sn+2 + 20/3)*9/4
Sn = 9/4Sn+2 + 15
C is the answer

let u swrite the $S_N$= $\frac{2}{3}$ $S_{N–1}$ – 4 in terms of N+2....
$S_{N+2}$= $\frac{2}{3}$ $S_{N+1}$ – 4, but $S_{N+1}$= $\frac{2}{3}$ $S_{N}$ – 4, so substitute this value in the previous equation..

$S_{N+2}$= $\frac{2}{3}$ ($\frac{2}{3}$ $S_{N}$ – 4) – 4 =>$S_{N+2}$= $\frac{2*2}{3*3}$ $S_{N}-\frac{2*4}{3}$ – 4..
=> $S_{N+2}$= $\frac{4}{9}$ $S_{N+1}-\frac{8}{3}$ – 4,
Multiply the equation by 9..
$9S_{N+2}$= 4 $S_{N}$-8*3 –9* 4 => 4 $S_{N}=9S_{N+2}$+60
Divide the entire equation by 4 to get value of $S_N$
$S_{N}=\frac{9}{4}S_{N+2}$+15

C

Test by numbers. Create a sequence from Sn to Sn+2 and then plug in Sn and Sn+2 values in the options and see if LHS = RHS.

Let Sn = 0.
Sn+1 = 2/3(0) - 4 = -4
Sn+2 = 2/3(-4) - 4 = -20/3.

Plug 0 and -20/3 on LHS and RHS of each option and choose that which matches.

We know that $S_N$= $\frac{2}{3}$ $S_{N–1}$ $- 4$

And we need to express $S_N$ in terms of $S_{N+2}$

Let's first write $S_{N+2}$
$S_{N+2}$ Can be expressed in terms of the previous term $S_{N+1}$ as
$S_{N+2}$= $\frac{2}{3}$ $S_{N+1}$ $- 4$ ...(1)

Similarly, $S_{N+1}$ Can be expressed in terms of the previous term $S_N$ as
$S_{N+1}$= $\frac{2}{3}$ $S_N$ $- 4$

Substituting Value of $S_{N+1}$ in ...(1) we get
$S_{N+2}$= $\frac{2}{3}$*( $\frac{2}{3}$ $S_N$ - 4) - 4
= $\frac{4}{9}$*$S_N$ - $\frac{8}{3}$ - 4
= $\frac{4}{9}$*$S_N$ - $\frac{20}{3}$
=> $\frac{4}{9}$*$S_N$ = $S_{N+2}$ + $\frac{20}{3}$
=> $S_N$ = $\frac{9}{4}$ * $S_{N+2}$ + $\frac{9*20}{(4*3)}$
=> $S_N$ = $\frac{9}{4}$ * $S_{N+2}$ + 15

So, answer will be C
Hope it helps!

To learn more about Sequences (Arithmetic and Geometric Sequence) watch the following video

Solution:

Lets say $S_N_-_1$=0
We can find the further values by substituting the above in the given function

$S_N$=-4
$S_N_+_1$=$\frac{-20}{3}$
$S_N_+_2$=$\frac{ -76}{9}$
Now, we know $S_N$=-4 and thus it is not fraction and therefore we need to remove the fraction in $S_N_+_2$
In order to remove he denominator we need to multiply the same number in the numerator. Thus, we look for choices having 9 in the numerator.

Eliminating choice B, D & E

We can then substitute the value of $S_N_+_2$ in option A to get the value of $S_N_$=-1
And by substituting in option C we get the value of $S_N_$= -4

IMO C

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.