# Given that the sum of n terms of a series is `(a^(2-n)(b^n - a^n))/(b-a)` , find the nth term.

### 2 Answers | Add Yours

a^(2-n)(b^n-a^n)/(b-a)

= a^2*a^(-n)(b^n-a^n)/(b-a)

= a^2(b^n*a^(-n)-a^n*a^(-n))/(b-a)

= a^2[(b/a)^n-a^(n-n)]/(b-a)

= a^2[(b/a)^(n-1)]/[a(b/a-1)]

= (a^2)/a[(b/a)^(n-1)]/(b/a-1)

`= a[(b/a)^(n-1)]/[(b/a)-1]`

We know for a geometric series;

Sn = a(r^n-1)/(r-1)

Where Sn is the sum of the series for n numbers and r is the common ratio and a is the first term.

Our question is also like a geometric series where;

`Sn = a^(2-n)(b^n-a^n)/(b-a) = a[(b/a)^(n-1)]/[(b/a)-1]`

`a=a`

`r=(b/a)`

So in this kind of series the nth term is given by;

`T_n = a*r^(n-1)`

So for our equation;

`T_n = a*(b/a)^(n-1) = (a^n)(b^(n-1))`

**Sources:**

The sum of n terms of a series is given as `S_n = (a^(2-n)(b^n - a^n))/(b-a).`

The nth term of the series is `T_n = S_n - S_(n - 1)`

= `(a^(2-n)(b^n - a^n))/(b-a) - (a^(2-(n - 1))(b^(n - 1) - a^(n - 1)))/(b-a)`

=> `(a^(2-n)(b^n - a^n) - a^(2-n + 1)(b^(n - 1) - a^(n - 1)))/(b-a)`

=> `(a^(2-n)/(b-a))*((b^n - a^n) - a*(b^(n - 1) - a^(n - 1)))`

**The nth term of the series is `T_n = (a^(2-n)/(b-a))*((b^n - a^n) - a*(b^(n - 1) - a^(n - 1)))`**