Arithmetic-geometric progression — finding sum: Question bank

Before the questions, three plain-word reminders so every symbol below is already earned:

True or false — justify

Recall Cover the answers — force a reason, not a guess

An AGP with $d = 0$ is just an ordinary GP. ::: True. With $d = 0$ every term is $a r^{k - 1}$ , the arithmetic part is frozen at $a$ , so it collapses to a pure GP. An AGP with $r = 1$ is just an ordinary AP. ::: True. With $r = 1$ the geometric factor $r^{k - 1} = 1$ always, leaving $\sum (a + (k - 1) d)$ , a plain AP sum — and this is exactly why the boxed formula (with $1 - r$ on the bottom) breaks and you must use the AP formula instead. The infinite sum $S_{\infty}$ exists for every AGP. ::: False. It exists only when $∣ r ∣ < 1$ ; otherwise the geometric part doesn't shrink, the terms don't die, and the total blows up. If $∣ r ∣ < 1$ then the AGP terms themselves always decrease from the start. ::: False. Early terms can grow because the arithmetic part $(a + (k - 1) d)$ is still rising; the terms only eventually shrink once the decaying $r^{k - 1}$ wins over the linear growth (see Limits: exponential vs polynomial growth). The arithmetic part being "linear" is why the correction term has $(1 - r)^{2}$ , not $(1 - r)^{3}$ . ::: True. A degree-1 (linear) coefficient needs the shift-and-subtract trick applied once, adding one extra factor of $(1 - r)$ beyond the GP's single $(1 - r)$ ; a quadratic coefficient would need it twice, giving $(1 - r)^{3}$ . Multiplying the series by $r$ and subtracting is the only correct way to derive $S_{n}$ . ::: False. It's the standard trick, but differentiating the GP identity $\sum r^{k}$ or using generating functions (Power series and generating functions) gives the same result — the multiply-and-shift is just the most elementary route. $\sum_{n \geq 1} n x^{n - 1} = \frac{1}{( 1 - x ) ^{2}}$ holds for all real $x$ . ::: False. Only for $∣ x ∣ < 1$ ; that's the AGP with $a = 1, d = 1, r = x$ , and outside the disk the series diverges even though the tidy formula $\frac{1}{( 1 - x ) ^{2}}$ still evaluates to a number.

Spot the error

Recall Each line has a hidden mistake — name it precisely

"For $\sum n \cdot 3^{n - 1}$ , use $S_{\infty} = \frac{1}{1 - 3} + \frac{3}{( 1 - 3 ) ^{2}} = - \frac{1}{2} + \frac{3}{4} = \frac{1}{4}$ ." ::: Error: $r = 3$ so $∣ r ∣ \neq < 1$ ; the series diverges and $S_{\infty}$ is meaningless. Plugging into the infinite formula gives a finite-looking number that is pure nonsense. "After subtracting $r S_{n}$ from $S_{n}$ , the leftover pure GP has $n$ terms." ::: Error: it has $n - 1$ terms ( $d r$ up to $d r^{n - 1}$ ). The first term $a$ and the last shifted term $(a + (n - 1) d) r^{n}$ are pulled out and are not part of the GP. "The correction term is $\frac{d r}{1 - r}$ because summing a GP gives a single $(1 - r)$ ." ::: Error: the denominator must be squared, $\frac{d r}{( 1 - r ) ^{2}}$ . One $(1 - r)$ comes from summing the leftover GP, the second from the $(1 - r)$ you factored out of $(1 - r) S_{n}$ . "In the derivation we set $a = 0$ to simplify, then add it back later." ::: Error: you cannot drop $a$ ; the very first surviving term of the subtraction is $a$ (it has nothing above it to cancel). Setting $a = 0$ deletes a real leftover. "Since the finite formula has $(1 - r)$ on the bottom, it fails whenever $r > 1$ ." ::: Error: it fails only at $r = 1$ (division by zero). For any other $r$ , including $r = 3$ , the finite $S_{n}$ formula is perfectly valid — only the infinite one demands $∣ r ∣ < 1$ . " $r S_{n}$ shifts every term one place left, lining term $k$ of $S_{n}$ under term $k$ of $r S_{n}$ ." ::: Error: multiplying by $r$ shifts one place right (raises each power of $r$ ), so term $k$ of $r S_{n}$ sits under term $k + 1$ of $S_{n}$ . That off-by-one alignment is the whole point. "We combine $\frac{a}{1 - x} + \frac{x}{( 1 - x ) ^{2}}$ by giving them denominator $(1 - x)$ ." ::: Error: the common denominator is $(1 - x)^{2}$ ; you must scale the first fraction by $(1 - x)$ , giving $\frac{( 1 - x ) + x}{( 1 - x ) ^{2}} = \frac{1}{( 1 - x ) ^{2}}$ .

Why questions

Recall Explain the machinery, not just the result

Why does multiplying by $r$ and subtracting help at all? ::: It makes consecutive AP coefficients line up, and their difference is the constant $d$ ; that constant coefficient turns the hard mixed series into a plain GP we already know how to sum. Why does the arithmetic part $(a + (n - 1) d)$ not stop the infinite sum from converging? ::: Because it grows only linearly while $r^{n}$ decays exponentially when $∣ r ∣ < 1$ , and exponential decay always overwhelms linear growth (Limits: exponential vs polynomial growth), so each term still $\to 0$ fast enough. Why is $S_{\infty}$ split into "a GP heart plus a tilt"? ::: $\frac{a}{1 - r}$ is what you'd get from a pure GP with constant coefficient $a$ ; the $\frac{d r}{( 1 - r ) ^{2}}$ term is the extra weight added because the coefficient tilts upward by $d$ each step instead of staying flat. Why does the special case $d = 0$ recover Sum to infinity of GP? ::: With $d = 0$ the tilt term $\frac{d r}{( 1 - r ) ^{2}}$ vanishes, leaving just $\frac{a}{1 - r}$ , the ordinary sum-to-infinity of a GP. Why must we treat $r = 1$ as a separate case entirely? ::: Every step of the derivation divides by $1 - r$ ; at $r = 1$ that is division by zero, so the AGP degenerates into a pure AP and needs the AP sum $\frac{n}{2} [2 a + (n - 1) d]$ instead. Why does $(n - 1) r^{n} \to 0$ as $n \to \infty$ when $∣ r ∣ < 1$ , even though $n - 1$ grows? ::: The exponential factor $r^{n}$ shrinks so fast that even multiplying by the growing $n - 1$ can't rescue it; polynomial growth loses to exponential decay, so the product collapses to $0$ .

Edge cases

Recall The scenarios people forget to check

What is $S_{\infty}$ when $r = 0$ ? ::: Only the first term survives ( $r^{k - 1} = 0$ for $k \geq 2$ ), so $S_{\infty} = a$ . The formula agrees: $\frac{a}{1 - 0} + \frac{d \cdot 0}{1} = a$ . Is $∣ r ∣ < 1$ satisfied by negative ratios like $r = - \frac{1}{2}$ ? ::: Yes — $∣ r ∣ < 1$ means the magnitude is under $1$ , so $r = - \frac{1}{2}$ converges. The signs make the terms alternate, but $∣ r ∣^{n}$ still shrinks to $0$ . What happens to the finite formula as $r \to 1$ from either side? ::: Both $\frac{a}{1 - r}$ and the correction blow up individually, but they cancel; taking the limit carefully recovers the AP sum $na + \frac{n ( n - 1 )}{2} d$ — the formula is removably singular at $r = 1$ . Does an AGP with $a < 0$ or $d < 0$ change any of the machinery? ::: No. $a$ and $d$ are just constants carried through the algebra; the derivation and both boxed formulas hold for any real $a, d$ (only $r$ controls convergence). What is the sum of a single-term AGP ( $n = 1$ )? ::: Just $a$ . Check the finite formula: the GP bracket $d r (1 - r^{0}) = d r (1 - 1) = 0$ and the last term cancels the $\frac{a}{1 - r}$ excess, leaving $S_{1} = a$ . If $r = 1$ and $d = 0$ , what is the series? ::: A constant series $a + a + a + \dots$ ; finitely it sums to $na$ , and its infinite sum diverges (unless $a = 0$ ). Both degeneracies stack: it's neither a proper GP nor a tilted AGP. Can the infinite AGP sum ever be negative? ::: Yes — e.g. $a$ or $d$ negative, or $r$ negative making alternating cancellation dominate; convergence ( $∣ r ∣ < 1$ ) says nothing about the sign of the total.

Connections

Geometric Progression — sum — the leftover series after subtraction is exactly this.
Sum to infinity of GP — the $d = 0$ edge case above.
Arithmetic Progression — the $r = 1$ degenerate case.
Power series and generating functions — the $\frac{1}{( 1 - x ) ^{2}}$ identity and an alternative derivation.
Limits: exponential vs polynomial growth — why the tail vanishes for $∣ r ∣ < 1$ .