Question

How do you write the nth term rule for the sequence `5/2,11/6,7/6,1/2,-1/6,...`?

Answer 1

`n^(th)` term of the series is `(19-4n)/6`

Explanation:

In `{5/2,11/6,7/6,1/2,-1/6...}`, let us make denominator common. As GCD is `6`, we can write the series as

`{15/6,11/6,7/6,3/6,-1/6...}`.

We observe that numerators form the series

`{15,11,7,3,-1,...}` which form an arithmetic series with first term `a` as `15` and common difference `d` between a term annd its preceding term is `12-15=-4`. Hence, `n^th` term of this series is `a+(n-1)d` i.e. `15+(n-1)×(-4)=15-4n+4=19-4n`

Hence `n^(th)` term of the series is `(19-4n)/6`