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Mathematics · Ch 8 — Sequences and Series

Sum to n Terms of a G.P.

8.4.2

Sum to n Terms of a G.P.

Sum to nn Terms of a Geometric Progression

We now turn to the problem of adding the first nn terms of a geometric progression. Let the first term be aa and the common ratio be rr. The sum of the first nn terms is denoted by SnS_n.

Sn=a+ar+ar2+⋯+arn−1(1)S_n = a + ar + ar^2 + \dots + ar^{n-1} \qquad(1)

The value of SnS_n depends critically on whether rr equals 1 or not.

Case 1: r=1r = 1

When the common ratio is 1, every term in the progression is identical to the first term aa. The sum of nn such terms is simply nn times aa.

Sn=a+a+a+⋯+a(n terms)=naS_n = a + a + a + \dots + a \quad (n \text{ terms}) = na

This is the straightforward case. The interesting work lies ahead.

Case 2: r≠1r \neq 1

When rr is not 1, we use a clever trick. Multiply equation (1) by rr:

rSn=ar+ar2+ar3+⋯+arn−1+arn(2)rS_n = ar + ar^2 + ar^3 + \dots + ar^{n-1} + ar^n \qquad(2)

Now subtract equation (2) from equation (1). Notice how almost every term cancels:

Sn−rSn=(a+ar+ar2+⋯+arn−1)−(ar+ar2+ar3+⋯+arn)(1−r)Sn=a−arn(1−r)Sn=a(1−rn)\begin{aligned} S_n - rS_n &= (a + ar + ar^2 + \dots + ar^{n-1}) - (ar + ar^2 + ar^3 + \dots + ar^n) \\ (1 - r)S_n &= a - ar^n \\ (1 - r)S_n &= a(1 - r^n) \end{aligned}

Since r≠1r \neq 1, we can divide both sides by (1−r)(1 - r) to obtain the formula for the sum.

Sn=a(1−rn)1−r,r≠1S_n = \frac{a(1 - r^n)}{1 - r}, \quad r \neq 1

An equivalent form, obtained by multiplying numerator and denominator by −1-1, is also commonly used:

Sn=a(rn−1)r−1,r≠1S_n = \frac{a(r^n - 1)}{r - 1}, \quad r \neq 1 …