Geometric Series and the Sum to Infinity

Four geometric series have common ratios $r = -0.9$, $r = -1.5$, $r = -1$ and $r = 1$ respectively. Which one has a sum to infinity?

• $r = -0.9$

  Correct. $|-0.9| = 0.9 < 1$, so the terms alternate in sign while shrinking by 10% in size every step, and both the terms and the gap to the total head to zero. The condition is on the size of $r$, so a negative ratio is perfectly admissible.

• B$r = -1.5$

  This satisfies $r < 1$ but not $|r| < 1$ — and it is the modulus that matters. $|-1.5| = 1.5$, so each term is half again as large as the previous one in size; the terms grow without limit while flipping sign, and no total exists. Reading the condition as "$r$ less than one" rather than "the size of $r$ less than one" is exactly what admits this case wrongly.

• C$r = -1$

  This is the boundary case, and the boundary genuinely fails. With $r = -1$ the series is $a - a + a - a + \cdots$ and its partial sums alternate between $a$ and $0$ forever, approaching neither. That is why the specification writes $|r| < 1$ strictly rather than $|r| \leq 1$.

• D$r = 1$

  With $r = 1$ every term equals $a$, so the sum of $n$ terms is $na$, which runs away for any non-zero $a$. The formula does not merely fail to converge here — it never applied, since deriving it required dividing by $1 - r$.

Traps tested: Convergence condition read as r less than one · Convergence boundary wrongly included · Constant series assumed to converge

For the geometric series $45 + 15 + 5 + \cdots$, find the sum of all the terms after the first two.

• $7.5$

  Correct. $r = \frac{15}{45} = \frac{1}{3}$, and the terms after the first two form a geometric series with the same ratio but first term $5$: $\frac{5}{1-\frac{1}{3}} = \frac{5}{\frac{2}{3}} = 7.5$. Cross-check against the whole series: $\frac{45}{2/3} = 67.5$, and $67.5 - 45 - 15 = 7.5$. ✓

• B$67.5$

  This is the sum of the entire series — the question's "after the first two" was not applied, so $a$ was taken as 45 rather than 5. The $a$ in $\frac{a}{1-r}$ is always the first term of the sum being asked for, not of the sequence it was carved out of.

• C$22.5$

  This removes only the first term, taking $a = 15$: $\frac{15}{2/3} = 22.5$, which equals $67.5 - 45$. One term too few has been deleted — the question excludes two. Writing out the terms that remain ($5, \frac{5}{3}, \frac{5}{9}, \ldots$) before summing removes the ambiguity.

• D$3.75$

  This computes $\frac{5}{1+\frac{1}{3}}$, a sign slip in the denominator. The formula is $\frac{a}{1-r}$, and here $r$ is positive, so $1 - r = \frac{2}{3}$, not $\frac{4}{3}$. The $1 + r$ shape only ever appears as the result of $1 - r$ with a negative $r$ substituted correctly.

Traps tested: Sum to infinity first term misidentified · Terms excluded miscounted · Sign error in one minus r

A geometric series has first term $4$ and common ratio $3$. Find the sum of the first $7$ terms.

• $4372$

  Correct. $S_7 = \frac{4(3^7 - 1)}{3-1} = \frac{4(2186)}{2} = 4372$, matching the direct sum $4+12+36+108+324+972+2916$. Using the $\frac{a(r^n-1)}{r-1}$ form here keeps every quantity positive; the booklet's $\frac{a(1-r^n)}{1-r}$ form gives $\frac{4(-2186)}{-2}$ and the same answer, with two more minus signs to track.

• B$2916$

  This is $u_7 = 4 \times 3^6 = 2916$, the seventh term rather than the sum of the first seven. It is also, revealingly, larger than half the correct total — with $r = 3$ the last term dominates the sum, which makes this distractor feel plausible in a way it would not for a slowly-growing series.

• C$13120$

  This uses $r^8$ instead of $r^7$: $\frac{4(3^8-1)}{2} = 13120$, the sum of eight terms rather than seven. The exponent in the sum formula is the count of terms being added, and there is no $-1$ on it — that $-1$ belongs to $u_n = ar^{n-1}$, where the exponent counts steps instead.

• D$-4372$

  This has the right size and the wrong sign — the signature of using $\frac{a(1-r^n)}{1-r}$ with $r > 1$ and cancelling only one of the two negatives. $1 - 3^7$ is negative and $1 - 3$ is negative, and a negative divided by a negative is positive. A sum of positive terms cannot be negative, which makes this one self-detectable in a second.

Traps tested: Nth term given instead of sum · Exponent off by one in sum formula · Double negative mishandled in sum formula

In $S_n = \frac{a(1-r^n)}{1-r}$, which single feature of the expression decides whether a sum to infinity exists?

• What $r^n$ does as $n$ grows — it is the only part of the expression that depends on $n$ at all

  Correct, and it is worth noticing how little work this takes: $a$ and $1-r$ are fixed constants, so the entire $n$-dependence of $S_n$ sits in $r^n$. If $r^n \to 0$ the whole expression tends to $\frac{a}{1-r}$; if $r^n$ grows or oscillates, $S_n$ inherits that behaviour and no limit exists. The convergence condition is a statement about $r^n$, translated into a statement about $r$.

• BThe sign of $a$

  $a$ is a constant multiplier sitting outside everything that varies. Changing its sign flips the sum from positive to negative but cannot change whether the sum settles at all — $\frac{a}{1-r}$ is finite or not regardless of which side of zero $a$ is on.

• CWhether $n$ is even or odd

  Parity of $n$ affects the sign of $r^n$ when $r$ is negative, but not its size — and it is the size that decides convergence. For $r = -0.9$ the terms alternate in sign at every step and the series converges anyway; for $r = -1.5$ they alternate too, and it diverges. Parity is not the variable doing the work.

• DWhether $1-r$ is positive

  $1 - r$ is a fixed constant, not something that changes as $n$ grows, so it cannot decide a question about limiting behaviour. Its sign affects the sign of the answer, and its being non-zero is what makes the formula legal at all ($r \neq 1$) — but neither of those is convergence.

Traps tested: Convergence attributed to first term · Convergence attributed to parity · Convergence attributed to denominator sign

In the standard proof that $S_n = \frac{a(1-r^n)}{1-r}$, why is the series multiplied by $r$ before the two lines are subtracted?

• Because multiplying by $r$ turns each term into the next one, shifting the whole list along by one place, so the two lines share every term but two and everything in between cancels on subtraction

  Correct. That shift is the entire mechanism: the only survivors of the subtraction are the $a$ at the front of the first line and the $ar^n$ at the end of the second, leaving $S_n(1-r) = a(1-r^n)$ immediately. Without the multiplication there is no overlap, and with no overlap there is nothing to cancel.

• BSo that the negative terms cancel out

  A geometric series need contain no negative terms at all for this proof to work — it holds identically for $3 + 6 + 12 + 24$. The cancellation is between identical terms appearing in both lines, not between terms of opposite sign.

• CBecause reversing and adding, as in the arithmetic proof, would also work here but takes longer

  Reversing and adding does not work here at all — it is not merely slower. Reversed pairs of a geometric series do not have a constant sum: for $3, 6, 12, 24$ the pairs come out $27, 18, 18, 27$. Constant pair-sums are a consequence of a constant additive step, which a geometric series does not have.

• DTo convert the geometric series into an arithmetic one, which is easier to sum

  Multiplying a geometric series by $r$ produces another geometric series with the same ratio — it stays geometric throughout, and nothing in this proof involves an arithmetic series at any stage.

Traps tested: Cancellation purpose confused with sign · Arithmetic proof method assumed transferable · Arithmetic and geometric formulas confused

The first 8 terms of a sequence form an arithmetic series with first term $30$ and common difference $-2$. From the 8th term onwards, the terms form a geometric series with common ratio $\frac{1}{2}$. What is the sum of all the terms of the sequence?

• $200$

  Correct. The 8th term is $30 + 7(-2) = 16$ and the arithmetic sum is $S_8 = \frac{1}{2}(8)[60 + 7(-2)] = 4(46) = 184$. The terms after the 8th are $8, 4, 2, \ldots$ — geometric with first term $16 \times \frac{1}{2} = 8$ — summing to $\frac{8}{1/2} = 16$. Total $184 + 16 = 200$. The other legal route agrees: $S_7 = 168$, plus $\frac{16}{1/2} = 32$, gives 200.

• B$216$

  This counts the 8th term twice — once inside $S_8 = 184$ and again as the first term of a geometric series summing to $\frac{16}{1/2} = 32$. The shared term belongs to exactly one of the two sums, and the difference between this answer and the correct one is precisely one copy of it: $216 - 200 = 16$.

• C$184$

  This is the arithmetic part alone, with the infinite geometric tail dropped. The tail is infinite in length but finite in total — $8 + 4 + 2 + 1 + \cdots = 16$ — so it contributes a real, computable amount that must be added on.

• D$192$

  This adds only the first term of the geometric tail ($184 + 8$) rather than its sum to infinity. The tail has infinitely many terms; their total happens to be exactly double its first term here, since $\frac{a}{1-r} = \frac{a}{1/2} = 2a$ when $r = \frac{1}{2}$.

Traps tested: Overlap term double counted · Geometric tail omitted · Infinite tail replaced by single term