Arithmetico-Geometric Sequence

An arithmetico-geometric sequence multiplies an arithmetic sequence and a geometric sequence together, term by term — most often the positions $1,2,3,\ldots$ multiplied by successive powers of a common ratio. Neither the arithmetic sum formula nor the geometric sum formula applies directly to such a sum, but the shift-and-subtract idea behind the geometric sum formula extends to handle it completely.

What you will learn

• Recognize a sequence formed by multiplying arithmetic and geometric terms
• Sum an arithmetico-geometric series using the shift-and-subtract technique
• Apply the technique to a contest-style sum

Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.

Recognizing the pattern

An arithmetico-geometric sequence looks like $1,\ 2r,\ 3r^2,\ 4r^3,\ldots$: the $k$th term is the $k$th term of an arithmetic sequence (here, just the position $k$) multiplied by the $k$th term of a geometric sequence (here, $r^{k-1}$). Sums of such sequences show up whenever a growing count is weighted by a shrinking or growing factor, and they cannot be summed with either the plain arithmetic or plain geometric sum formula alone, since neither formula accounts for both patterns at once.

The telltale sign is a sum where the coefficients increase (or otherwise follow an arithmetic pattern) while the powers of some fixed number increase alongside them, as in $S=1+2r+3r^2+4r^3$. Spotting this shape is the first step; the shift-and-subtract technique below is what actually evaluates it.

The shift-and-subtract technique

The same idea that derives the geometric sum formula solves this harder case: multiply the entire sum by the common ratio, then subtract. For $S=1+2r+3r^2+4r^3$, multiplying every term by $r$ shifts each power up by one: $rS=r+2r^2+3r^3+4r^4$. Subtracting lines up matching powers of $r$: $S-rS=1+(2-1)r+(3-2)r^2+(4-3)r^3-4r^4=1+r+r^2+r^3-4r^4$.

The coefficients that survive the subtraction are all 1, so what remains on the right side is a plain geometric series, $1+r+r^2+r^3$, minus one leftover boundary term, $4r^4$, coming from the highest-indexed term that had nothing to subtract against. Collecting everything gives $(1-r)S=(1+r+r^2+r^3)-4r^4$, and dividing both sides by $1-r$ (valid whenever $r\neq1$) isolates $S$.

Scaling the technique to a longer sum

The same three moves — multiply by $r$, subtract, then divide by $1-r$ — apply no matter how many terms the sum has. For $n$ terms, the general pattern is $(1-r)S=(1+r+r^2+\cdots+r^{n-1})-nr^n$: the geometric part always runs from $r^0$ to $r^{n-1}$, and the boundary term is always $n$ (the last coefficient) times $r^n$ (one power beyond the geometric part).

This means the geometric part can be evaluated with the ordinary geometric sum formula from the previous lesson, leaving only careful bookkeeping of the single boundary term and one final division. The technique scales to long sums exactly because it replaces $n$ separate additions with one geometric-sum calculation and one subtraction.

See the reasoning unfold

Pause after each step and predict what comes next.

Common mistakes to catch

• Trying to apply the plain geometric sum formula directly to a sum whose coefficients are not all equal to 1.
• Forgetting the leftover boundary term (the highest coefficient times $r^n$) that remains after subtracting $rS$ from $S$.
• Dividing by $1-r$ before finishing the subtraction, instead of first simplifying the right-hand side completely.

Try with support

Guided practice

Choose a strategy, take a first step, and use the hints when you need direction.

Put it into practice

Independent practice

Work on your own first. If you reveal a solution, work through it and mark it reviewed.

Bring it together

• An arithmetico-geometric sum multiplies an arithmetic sequence (often just the positions $1,2,3,\ldots$) by a geometric sequence, term by term.
• Multiplying the sum $S$ by the common ratio $r$ and subtracting collapses the coefficients to a plain geometric series, leaving one leftover boundary term.
• Divide the simplified result by $1-r$ to isolate $S$; the geometric part can always be evaluated with the standard geometric sum formula.