(Sequence), Sn = 5, S2n = 650, S3n = ?

Question

A questions using geometric sequence.

Answer 1

S_n = sum of first n terms  = a₁(1-rⁿ)/(1-r) = 5

 

S_2n =sum of first 2n terms = a₁(1-r²ⁿ)/(1-r) = 650 = 5(130)

 

So, 130a₁(1-rⁿ)/(1-r) = a₁(1-r²ⁿ)/(1-r)

 

      130(1-rⁿ) = 1-r²ⁿ

 

      r²ⁿ - 130rⁿ + 129 = 0

 

      Let u = rⁿ  Then, u² - 130u + 129 = 0

 

                               (u-129)(u-1) = 0  So, u = 129 or u = 1

 

That is, rⁿ = 129 or rⁿ = 1.    

 

If rⁿ =1,  then the formula for S_n  gives S_n = 0, not 5.

 

Therefore, rⁿ = 129

 

Since S_n = a₁(1-rⁿ) / (1-r) = 5, we get -128a₁ = 5 - 5r

 

        So, a₁ = (5r-5)/128

 

Thus, S_3n = a₁(1-r³ⁿ)/(1-r)

 

               = [5(r-1)/128] (1 - (rⁿ)³] / (1-r)

 

               = (-5/128)[1 - (129)³]

 

               = 83855

 

        

Answer 2

If you are saying that the nth number of a geometric sequence is 5 and the 2n^th term equals 650, I infer that to get from the nth term to the 2n^th term there must be n multiplications by a constant multiplier r.

 

5(rⁿ) = 650

rⁿ = 130

 

To get from the 2n^th term to the 3n^th term, multiply 650 by rⁿ.  Therefore S_3n = 84,500