In short

An arithmetico-geometric series (AGS) is formed when the k-th term is the product of the k-th term of an AP and the k-th term of a GP: T_k = [a + (k-1)d]\,r^{k-1}. To sum it, multiply the entire sum by r, shift, and subtract — the same trick that works for a pure GP, except now each subtraction leaves behind a GP of its own. The finite sum is S_n = \dfrac{a - [a+(n-1)d]\,r^n}{1-r} + \dfrac{dr(1-r^{n-1})}{(1-r)^2} for r \neq 1, and the infinite sum (when |r|<1) simplifies to S_\infty = \dfrac{a}{1-r} + \dfrac{dr}{(1-r)^2}.

You start a new savings plan. In the first month you deposit ₹1,000. Each following month, you increase the deposit by ₹500 (so the second month is ₹1,500, the third is ₹2,000, and so on). The bank pays monthly interest at a rate that effectively multiplies each past deposit by 1.01 every month. After 12 months, how much money is in the account?

The deposit in month k is 1000 + (k-1) \cdot 500 — an arithmetic progression. Each deposit compounds by a factor of 1.01 for every month it sits in the account, so the contribution of month k to the final total is [1000 + (k-1) \cdot 500] \cdot (1.01)^{12-k} — an AP term multiplied by a GP term. That product is the signature of an arithmetico-geometric series.

You already know how to sum a pure GP (multiply by r, subtract, and collapse). You already know how to sum a pure AP (pair the first and last terms). But when every term is an AP value times a GP value, neither trick works alone. The AGS formula handles exactly this hybrid.

What makes a series arithmetico-geometric

Take an AP a, a+d, a+2d, \dots and a GP 1, r, r^2, \dots and multiply them term by term:

a \cdot 1,\quad (a+d) \cdot r,\quad (a+2d) \cdot r^2,\quad (a+3d) \cdot r^3,\quad \dots

The general term is

T_k = [a + (k-1)d]\,r^{k-1}, \qquad k = 1, 2, 3, \dots

The AP part (a + (k-1)d) grows linearly. The GP part (r^{k-1}) grows (or shrinks) exponentially. Their product creates a sequence that neither grows purely linearly nor purely exponentially — it blends both behaviours.

AP terms, GP terms, and their product forming an AGSThree rows of boxes. Row 1 (AP): 2, 5, 8, 11, 14. Row 2 (GP): 1, 1/2, 1/4, 1/8, 1/16. Row 3 (AGS = AP times GP): 2, 5/2, 2, 11/8, 14/16. Multiplication signs connect corresponding boxes in rows 1 and 2 to row 3. AP: 2 5 8 11 14 GP: 1 1/2 1/4 1/8 1/16 × × × × × AGS: 2 5/2 2 11/8 7/8 AP × GP = AGS — each term blends linear growth with exponential scaling.
An arithmetico-geometric series is built by multiplying an AP ($2, 5, 8, 11, 14$ with $a=2$, $d=3$) and a GP ($1, \tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{8}, \tfrac{1}{16}$ with $r = \tfrac{1}{2}$) term by term. The AGS terms (bottom row) are $2, \tfrac{5}{2}, 2, \tfrac{11}{8}, \tfrac{7}{8}$.

Arithmetico-Geometric Series

An arithmetico-geometric series is a series whose k-th term is the product of the k-th term of an AP and the k-th term of a GP:

T_k = [a + (k-1)d]\,r^{k-1}

where a is the first term of the AP, d is the common difference, and r is the common ratio of the GP.

The sum of the first n terms (for r \neq 1) is:

S_n = \frac{a - [a + (n-1)d]\,r^n}{1-r} + \frac{dr(1 - r^{n-1})}{(1-r)^2}

When |r| < 1, the infinite sum is:

S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2}

Deriving the sum formula

The derivation uses the same multiply-shift-subtract trick that works for a pure GP, but with a twist: the subtraction does not collapse everything into two terms. It leaves behind a pure GP, which you then sum separately.

Step-by-step derivation

Write out the sum:

S_n = a + (a+d)r + (a+2d)r^2 + (a+3d)r^3 + \cdots + [a+(n-1)d]\,r^{n-1}

Multiply both sides by r:

rS_n = ar + (a+d)r^2 + (a+2d)r^3 + (a+3d)r^4 + \cdots + [a+(n-1)d]\,r^n

Subtract the second equation from the first:

S_n - rS_n = a + dr + dr^2 + dr^3 + \cdots + dr^{n-1} - [a+(n-1)d]\,r^n

Look at the right-hand side. The first surviving term is a. The middle terms are dr + dr^2 + \cdots + dr^{n-1} — this is a pure GP with first term dr, common ratio r, and n-1 terms. The last term is -[a+(n-1)d]\,r^n.

The multiply-shift-subtract derivation of the AGS sum formulaTwo rows of terms. Row 1 is S sub n with terms a, (a+d)r, (a+2d)r squared, and so on. Row 2 is rS sub n, shifted one position right, with terms ar, (a+d)r squared, (a+2d)r cubed, and so on. Subtraction yields a at the left, d times r plus d times r squared plus dots in the middle (a GP), and minus the last term of rS at the right. S n = a (a+d)r (a+2d)r² ··· [a+(n−1)d]rⁿ⁻¹ rS n = ar (a+d)r² ··· [a+(n−2)d]rⁿ⁻¹ [a+(n−1)d]rⁿ diff = dr diff = dr² subtract: a + dr + dr² + ··· + drⁿ⁻¹ − [a+(n−1)d]rⁿ ↑ this is a pure GP with n−1 terms (1−r)Sₙ = a + dr·(1−rⁿ⁻¹)/(1−r) − [a+(n−1)d]rⁿ
The multiply-shift-subtract trick applied to an AGS. Unlike a pure GP where everything cancels except two terms, the subtraction here leaves a GP ($dr + dr^2 + \cdots + dr^{n-1}$) in the middle. That GP is then summed using the standard GP formula.

Factor the left side and sum the GP in the middle:

(1 - r)\,S_n = a + dr \cdot \frac{1 - r^{n-1}}{1 - r} - [a + (n-1)d]\,r^n

Divide both sides by (1 - r):

\boxed{S_n = \frac{a - [a + (n-1)d]\,r^n}{1 - r} + \frac{dr(1 - r^{n-1})}{(1 - r)^2}}

The formula has two pieces: the first fraction handles the "endpoints" (a at the start, the last AGS term at the end), and the second fraction handles the GP that the subtraction produced. When d = 0, the AGS reduces to a pure GP, and the second fraction vanishes — the formula collapses to S_n = a \cdot \frac{1 - r^n}{1 - r}, exactly the GP sum.

The infinite sum

When |r| < 1, the terms r^n and r^{n-1} both tend to 0 as n \to \infty. The term [a + (n-1)d]\,r^n also tends to 0 because the linear growth of a + (n-1)d is overwhelmed by the exponential decay of r^n. The finite sum formula simplifies:

\boxed{S_\infty = \frac{a}{1 - r} + \frac{dr}{(1 - r)^2}}

This formula is compact enough to memorise and appears frequently in JEE problems.

Partial sums of the AGS with a=1, d=1, r=1/2 approaching 4A graph with n on the horizontal axis from 1 to 10 and partial sum on the vertical axis from 0 to 5. A dashed red line at height 4 marks the infinite sum. Ten dots show the partial sums climbing from 1 toward 4, getting closer with each term. n Sₙ 0 1 2 3 4 S∞ = 4 1 2 3 4 5 6 7 8 9 1 2.5 3.25 3.625
The partial sums of the AGS $1 + \tfrac{2}{2} + \tfrac{3}{4} + \tfrac{4}{8} + \cdots$ (where $a = 1$, $d = 1$, $r = \tfrac{1}{2}$) climb toward $S_\infty = \tfrac{1}{1 - 1/2} + \tfrac{1 \cdot 1/2}{(1 - 1/2)^2} = 2 + 2 = 4$. By the 5th partial sum, you are already within $3\%$ of the limit.

A quick sanity check

For a = 1, d = 1, r = \frac{1}{2}, the series is 1 + \frac{2}{2} + \frac{3}{4} + \frac{4}{8} + \frac{5}{16} + \cdots

Partial sums: S_1 = 1, S_2 = 2, S_3 = 2.75, S_4 = 3.25, S_5 = 3.5625. These are climbing toward the predicted S_\infty = 4, confirming the formula.

Applications

Sums involving k \cdot r^k

The most frequently tested AGS in JEE is \sum_{k=1}^{n} k \, x^k, which is simply S_n with a = 1, d = 1. Using the infinite sum formula with |x| < 1:

\sum_{k=1}^{\infty} k\,x^k = \frac{x}{(1-x)^2}

This result also follows from differentiating the geometric series \sum_{k=0}^{\infty} x^k = \frac{1}{1-x} with respect to x and then multiplying by x. The AGS formula and the calculus approach give the same answer — a reassuring consistency.

Weighted averages with decaying weights

Suppose you are computing a weighted average of the numbers 1, 2, 3, 4, \dots where the weight of the k-th number is r^{k-1} (with 0 < r < 1). The numerator is \sum_{k=1}^{\infty} k \cdot r^{k-1} = \frac{1}{(1-r)^2} (an AGS with a = 1, d = 1, each term divided by r compared to the standard form). The denominator is \sum_{k=1}^{\infty} r^{k-1} = \frac{1}{1-r}. The weighted average is \frac{1}{1-r} — a clean result from the AGS formula.

Decaying weights on terms 1 through 6Six vertical bars of decreasing height. The k-th bar represents the weight r to the power k minus 1 applied to the value k. Bar heights decrease geometrically. Labels show the product k times r to the k minus 1 for each bar. k=1 1 k=2 2r k=3 3r² k=4 4r³ k=5 5r⁴ ···
Each term $k \cdot r^{k-1}$ (shown here for $r = \tfrac{1}{2}$) is the product of a linearly growing factor ($k$) and an exponentially decaying weight ($r^{k-1}$). The bars grow taller due to $k$ but shrink due to $r^{k-1}$, with the decay eventually winning. The sum of all bar heights is $\frac{1}{(1-r)^2}$.

Two worked examples

Example 1: Find the sum $1 \cdot 1 + 2 \cdot \frac{1}{3} + 3 \cdot \frac{1}{9} + 4 \cdot \frac{1}{27} + \cdots$ to infinity

Step 1. Identify the AP and GP components.

T_k = k \cdot \left(\frac{1}{3}\right)^{k-1}

So a = 1, d = 1, r = \frac{1}{3}.

Why: each term has the form (AP term) × (GP term). The AP is 1, 2, 3, 4, \dots and the GP is 1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots

Step 2. Verify that |r| < 1.

|r| = \frac{1}{3} < 1 \quad \checkmark

Why: the infinite sum formula only applies when the GP component decays, which requires |r| < 1.

Step 3. Apply the infinite sum formula.

S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = \frac{1}{1 - 1/3} + \frac{1 \cdot (1/3)}{(1 - 1/3)^2}
= \frac{1}{2/3} + \frac{1/3}{4/9} = \frac{3}{2} + \frac{1}{3} \cdot \frac{9}{4} = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}

Why: the first fraction (\frac{3}{2}) is the GP contribution, and the second (\frac{3}{4}) captures the extra weight from the AP factor growing linearly.

Step 4. Verify with partial sums. S_1 = 1. S_2 = 1 + \frac{2}{3} = \frac{5}{3} \approx 1.667. S_3 = \frac{5}{3} + \frac{3}{9} = 2. S_4 = 2 + \frac{4}{27} \approx 2.148. These are climbing toward \frac{9}{4} = 2.25.

Result: S_\infty = \dfrac{9}{4} = 2.25.

Partial sums of AGS with a=1 d=1 r=1/3 approaching 9/4A graph with n on the horizontal axis from 1 to 8 and partial sum on the vertical axis from 0 to 2.5. A dashed red line at 2.25 marks the infinite sum. Eight dots climb from 1 toward the red line. 0 1 2 S∞ = 9/4 1 2 3 4 5 6 7 1 5/3 2
The partial sums of $1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \cdots$ climb toward $\frac{9}{4} = 2.25$. By $n = 4$ the sum is already $2.148$, within $5\%$ of the limit. The convergence is fast because $r = \frac{1}{3}$ is well below $1$.

The graph confirms what the formula predicts: even though the AP factor grows (1, 2, 3, 4, \dots), the GP factor (1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots) decays fast enough that the sum converges to a finite limit.

Example 2: Find the sum of the first 5 terms of the AGS $3 + 5 \cdot 2 + 7 \cdot 4 + 9 \cdot 8 + 11 \cdot 16$

Step 1. Identify a, d, r, and n.

The AP is 3, 5, 7, 9, 11, so a = 3, d = 2. The GP is 1, 2, 4, 8, 16, so r = 2. And n = 5.

Why: the AP terms increase by 2 each time, and the GP terms double each time.

Step 2. Use the multiply-and-subtract method directly (since r = 2 makes the formula's denominator simple).

S_5 = 3 + 10 + 28 + 72 + 176
2S_5 = 6 + 20 + 56 + 144 + 352
S_5 - 2S_5 = 3 + (10-6) + (28-20) + (72-56) + (176-144) - 352
-S_5 = 3 + 4 + 8 + 16 + 32 - 352

Why: each difference in the middle is d \cdot r^{k-1} = 2 \cdot 2^{k-1}, forming a GP.

Step 3. Sum the GP 4 + 8 + 16 + 32 (first term 4, ratio 2, 4 terms).

4 \cdot \frac{2^4 - 1}{2 - 1} = 4 \cdot 15 = 60

Why: this is the standard GP sum with a = 4, r = 2, n = 4.

Step 4. Combine.

-S_5 = 3 + 60 - 352 = -289
S_5 = 289

Step 5. Verify by direct addition: 3 + 10 + 28 + 72 + 176 = 289. Confirmed.

Result: S_5 = 289.

Five bars showing the AGS terms summing to 289Five horizontal bars stacked vertically, representing the five terms: 3, 10, 28, 72, 176. Each bar is wider than the previous. The total 289 is shown below. 3 10 28 72 176 3 + 10 + 28 + 72 + 176 = 289
The five AGS terms grow rapidly because $r = 2 > 1$: the GP factor doubles each time while the AP factor adds $2$. The last term alone ($176$) is more than $60\%$ of the total, just as with a pure GP where $r > 1$.

When r > 1, the AGS diverges — the terms grow without bound, so there is no infinite sum. The finite sum formula still works, and this example shows the multiply-subtract method applied step by step with concrete numbers.

Common confusions

Going deeper

If you came here to learn what an AGS is, how to derive its sum, and how to apply the formula, you have everything you need. The rest of this section is for readers who want to see the technique generalised and connected to other ideas.

The multiply-subtract trick is more general than the AGS

The derivation multiplied by r and subtracted to reduce the "AP × GP" to a "pure GP." The same strategy works when the AP is replaced by a quadratic sequence (T_k = (ak^2 + bk + c) \cdot r^{k-1}) — you just need to apply the trick twice. The first subtraction replaces the quadratic coefficient with a linear one (creating an AGS), and the second subtraction reduces the AGS to a GP. In general, if the non-geometric factor is a polynomial of degree m, you apply the multiply-subtract trick m + 1 times.

Connection to differentiation of power series

There is a slick alternative derivation of \sum_{k=1}^{\infty} k\,x^{k-1}. Start with the geometric series:

\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}, \qquad |x| < 1

Differentiate both sides with respect to x:

\sum_{k=1}^{\infty} k\,x^{k-1} = \frac{1}{(1-x)^2}

Multiply both sides by x:

\sum_{k=1}^{\infty} k\,x^k = \frac{x}{(1-x)^2}

This matches the AGS formula with a = 1, d = 1. Differentiating again and multiplying by x gives \sum k^2 x^k, and so on. Each differentiation "promotes" the power of k by one, generating AGS-like sums with higher polynomial coefficients.

Differentiation of the geometric series producing AGS sumsThree rows connected by downward arrows labelled differentiate and multiply by x. Row 1: sum of x to the k equals 1 over 1 minus x. Row 2: sum of k times x to the k minus 1 equals 1 over (1 minus x) squared. Row 3: sum of k times x to the k equals x over (1 minus x) squared. Σ xᵏ = 1/(1−x) differentiate Σ k·xᵏ⁻¹ = 1/(1−x)² multiply by x Σ k·xᵏ = x/(1−x)² Same result as the AGS formula with a=1, d=1.
Differentiating the geometric series and multiplying by $x$ produces the AGS sum $\sum k x^k = \frac{x}{(1-x)^2}$. This technique generates closed forms for $\sum k^m x^k$ for any positive integer $m$ by repeated differentiation.

A general viewpoint

The AGS formula is a special case of a broader principle: whenever you can reduce a complicated series to a simpler one by a shift-and-subtract operation, do it. The multiply-subtract trick for GPs, the AGS derivation, and the Method of Differences are all instances of this principle. The idea reappears in generating functions, recurrence relations, and the calculus of finite differences — different languages for the same core move.

Where this leads next