![]() The sum of GP (of infinite terms) is: S ∞ = a/(1-r), when |r| The sum of GP (of n terms) is: S n = na, when r = 1.The sum of GP (of n terms) is: S n = a(r n - 1) / (r - 1) S n = a(1 - r n) / (1 - r), if r ≠ 1.is an infinite geometric series with a = 1 and r = 1/2. Taking 's 2' as common factor, the sum of areas is, s 2 ( 1 + 1/2 + 1/4 +. The areas of squares thus formed are, s, s 2/2, s 2/4, s 2/8. If a side of the first square is "s" units, determine the sum of areas of all the squares so formed? A third square is drawn inside the second square in the same way and this process is continued indefinitely. If |r| ≥ 1, then the geometric series diverges and it cannot have a sum.Įxample: A square is drawn by joining the midpoints of the sides of the original square. ![]() If |r| Note that for any infinite geometric series, This is the formula for the sum of infinite GP. Let us assume the sum of all these infinite number of terms be S ∞. (up to an infinite number of terms) such that the absolute value of its common ratio is less than 1. But what if we have to find the sum of an infinite GP? Let us consider a GP a, ar, ar 2. The above formula gives the sum of a finite GP. We have got the same answer using the GP sum formula also. Here, the first term is, a = 2, the common ratio is, r = 2, and the number of terms is, n = 6 (as we want the sum of the amounts after 6 weeks). Clearly, this is a geometric progression as 4/2 = 8/4 = 16/2 =. The amounts saved by Clara in the order of weeks are, 2, 4, 8, 16. Now, let us work on the example (from the last section) using the sum of n terms of GP formula. What happens when r = 1? Then the GP is of form a, a, a. Subtracting equation (1) from equation (2): Let S be the sum of the geometric progression of n terms. Then the first 'n' terms of GP are of the form a, ar, ar 2. Consider the sum of the first n terms of a geometric progression (GP) with first term a and common ratio r. ![]() Let's discuss how to calculate the sum of n terms of GP.
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