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

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