What is the sum of n terms of a geometric series?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The terms of a geometric progression have a common ratio. If the common ratio is r and the first term is a, the nth term of a geometric series is a*r^(n - 1)

The sum of n terms of a geometric progression is:

`sum_(k=0)^n( Tk)` = a + a*r + a*r^2 + ... a*r^n

=> (1 - r)`sum_(k=0)^n (Tk)` => (1 - r)(a + a*r + a*r^2 + ... a*r^(n -1)

=> (1 - r)`sum_(k = 0)^n(Tk)` => a + ar + ar^2 + ... ar^n - ar - ar^2 - ... - ar^(n + 1)

=> (1 - r)`sum_(k = 0)^n (Tk)` => a - ar^(n + 1)

=> `sum_(k = 0)^n(Tk)` = (a - ar^(n+1)/(1 - r)

=> `sum_(k = 0)^n(Tk)` = a(1 - r^(n+1))/(1 - r)

The sum of n terms of a geometric progression is a(1 - r^(n+1))/(1 - r)

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