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Mathematics · Ch 10 — Sequence and Series

Arithmetic-Geometric Progression

7

Arithmetic-Geometric Progression

A series is called an Arithmetic-Geometric Progression (A.G.P.) if its nnth term is the product of the nnth term of an A.P. and the nnth term of a G.P. If the A.P. is a,a+d,a+2d,…a, a+d, a+2d,\ldots and the G.P. is 1,r,r2,…1, r, r^2, \ldots, the A.G.P. is

a, (a+d)r, (a+2d)r2, …, [a+(n−1)d]rn−1, …a,\ (a+d)r,\ (a+2d)r^2,\ \ldots,\ \big[a+(n-1)d\big]r^{n-1},\ \ldots

so its nnth term is tn=[a+(n−1)d]rn−1t_n = \big[a+(n-1)d\big]r^{n-1}. Such series arise naturally whenever a linearly-growing quantity is discounted or weighted by a geometrically-shrinking (or growing) factor.

Sum to nn terms — derivation. Let

S=a+(a+d)r+(a+2d)r2+⋯+[a+(n−1)d]rn−1,r≠1.S = a + (a+d)r + (a+2d)r^2 + \cdots + \big[a+(n-1)d\big]r^{n-1}, \qquad r\ne 1.

Following the G.P. strategy of Section 4, multiply throughout by rr and align terms one position to the right:

rS=ar+(a+d)r2+⋯+[a+(n−2)d]rn−1+[a+(n−1)d]rn.rS = ar + (a+d)r^2 + \cdots + \big[a+(n-2)d\big]r^{n-1} + \big[a+(n-1)d\big]r^n.

Subtracting, every "column" from arar through [a+(n−2)d]rn−1\big[a+(n-2)d\big]r^{n-1} appears in both lines, but unlike a pure G.P. the terms do not cancel completely — each pairing leaves behind exactly d rkd\,r^k, since [a+kd]rk−[a+(k−1)d]rk\big[a+kd\big]r^k - \big[a+(k-1)d\big]r^k's partner in the subtraction is what remains after aligning by shift. Carrying out the subtraction carefully term-by-term:

S−rS=a+(dr+dr2+⋯+drn−1)−[a+(n−1)d]rn.S - rS = a + \Big(dr+dr^2+\cdots+dr^{n-1}\Big) - \big[a+(n-1)d\big]r^n.

The bracketed middle piece is dd times a plain geometric series of (n−1)(n-1) terms, r+r2+⋯+rn−1=r(1−rn−1)1−rr+r^2+\cdots+r^{n-1} = \dfrac{r(1-r^{n-1})}{1-r} (Section 4's formula, first term rr, ratio rr, n−1n-1 terms), so

S(1−r)=a+dr(1−rn−1)1−r−[a+(n−1)d]rn.S(1-r) = a + \frac{dr(1-r^{n-1})}{1-r} - \big[a+(n-1)d\big]r^n.

Dividing throughout by (1−r)(1-r) gives the boxed result:

Sn=a1−r+dr(1−rn−1)(1−r)2−[a+(n−1)d]rn1−r(r≠1).\boxed{S_n = \frac{a}{1-r} + \frac{dr\big(1-r^{n-1}\big)}{(1-r)^2} - \frac{\big[a+(n-1)d\big]r^n}{1-r}} \qquad (r\ne 1).

Although the formula looks heavier than the pure A.P. or G.P. sums, it is built from exactly the same two ideas used in Sections 2 and 4 — reversing/shifting and subtracting — applied one after the other. …

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