Given the sum of a series and a term how would you find Tn?

[Superposed_Cat]

Hi all, given the sum of a series and a single term how would one find the nth term? Any help appreciated.

 

[HallsofIvy]

You can't. That's obviously impossible. There exist an infinite number of series having a given specific term and converging to a give value.

Suppose you are given the term "$a_i$", for i a specific number, and that the sum is "S". Choose "A" to be any number and write the geometric sequence that converges to $S- a_i$. Then the series consisting first term A, the rest of the geometric series with "$a_i$" stuck in a the ith place, has sum S.

 

[Superposed_Cat]

It's arithmetic, never mind I figured it out, apologies for wasted time.

t6=32
s9=234
32=a+5d
d=(32-a)/5

234=0.5*9(2a+8((32-a)/5))
a=2
d=6
Tn=6n-4

 

[HallsofIvy]

In your original post you said "given the sum of a series and a single term". What are "t6", "s9", "a" and "d".

 

[Superposed_Cat]

a is the first term, d is the difference between terms, t6 is 6th term, s9 is sum of the first 9 terms