Chapter 8 — Sequences and Series

Answer: A

Explanation

The common difference is the fixed amount added each time. Subtract any term from the next:

$$d = T_2 - T_1 = 11 - 7 = 4$$

Confirm on another pair: $15-11 = 4$ ✓ and $19-15 = 4$ ✓

Why the others are wrong:

• D (6) divided the span by the wrong gap count: $(19-7)/2$. Four terms have three gaps, so $(19-7)/3 = 4$.
• B (3) is the gap count, not the gap size.
• C (5) counted numbers instead of steps: $7,8,9,10,11$ is five numbers but four steps.

Takeaway: $d = T_{n+1} - T_n$, checked on two pairs. Never confuse "how many gaps" with "how big is each gap".

Answer: C

Explanation

Substitute into $T_n = a+(n-1)d$ and expand fully:

$$T_n = 3 + (n-1)(5) = 3 + 5n - 5 = 5n - 2$$

The one-second test: substitute $n=1$. The formula must give the first term, 3. Option C gives $5(1)-2 = 3$ ✓

Why the others are wrong — each fails that test:

• A ($5n+3$) gives $T_1 = 8$. This is $a+dn$ — the "forgot the $-1$" version.
• B ($5n$) gives $T_1 = 5$. The constant vanished.
• D ($3n+5$) gives $T_1 = 8$. The roles of $a$ and $d$ are swapped — the coefficient of $n$ must be $d$.

Takeaway: Every AP expands to $T_n = dn + (a-d)$. Whatever you derive, test it at $n=1$. Wrong formulas rarely survive one substitution.

Answer: B

Explanation

Read off $a = 2$ and $d = 3$, then think of the journey: to reach the 15th term you start at 2 and take fourteen steps of 3 (the first term needs no step).

$$T_{15} = 2 + (15-1)(3) = 2 + 42 = 44$$

Why the others are wrong: the options are a map of every way to fumble $a + (n-1)d$ —

• A (42) is $(15-1) \times 3$ alone — the steps were taken but the starting value was never added.
• C (47) is $2 + 15 \times 3$ — fifteen steps instead of fourteen: the "$n$ instead of $n-1$" slip.
• D (41) is $2 + 13 \times 3$ — an over-correction to thirteen steps ($n-2$), usually from "subtracting 1" twice.

Takeaway: $T_n = a + (n-1)d$: the $n$th term is reached after $n-1$ steps, never $n$. Say it as a story — "start at $a$, step $n-1$ times" — and the off-by-one options lose their pull.

Answer: D

Explanation

The value is known; the position is wanted. Set the general term equal to 100:

$$1 + (n-1)(3) = 100 \implies 3n = 102 \implies n = 34$$

Check forwards: $T_{34} = 1 + 33(3) = 100$ ✓

Why the others are wrong:

• B (33rd) solved to $n-1 = 33$ and stopped. That is the number of steps, not the position.
• C (35th) added 1 to a finished answer.
• A (32nd) counted from $T_2 = 4$ instead of the first term.

Takeaway: $n-1$ and $n$ are different numbers, and the wrong options wait for whichever you confuse. The forward check settles it in five seconds.

Answer: A

Explanation

In three consecutive AP terms, the middle one is the average of its neighbours. So double the middle term equals the sum of the outer two:

$$2(2k+1) = k + (4k-1) \implies 4k+2 = 5k-1 \implies k = 3$$

Check: the terms are $3, 7, 11$ — differences 4 and 4 ✓

Why the others are wrong: substitute each and the equal-differences test fails.

• B ($k=2$): $2, 5, 7$ — gaps 3 and 2 ✗
• C ($k=4$): $4, 9, 15$ — gaps 5 and 6 ✗
• D ($k=-3$): $-3, -5, -13$ — gaps $-2$ and $-8$ ✗

Takeaway: Three consecutive AP terms satisfy $2q = p+r$. Solve, then substitute back — if the differences are not equal, it is not the answer.

Section 8.2 — Arithmetic Sequences: More Problems

Answer: C

Explanation

Two non-consecutive terms still pin down $d$, because the distance between them is a known number of equal jumps. Position 3 to position 7 is $7-3 = 4$ jumps:

$$4d = 27 - 11 = 16 \implies d = 4$$

Check: $11 \to 15 \to 19 \to 23 \to 27$ ✓ — four hops of 4.

Why the others are wrong: each fails the same forward check.

• A ($d=2$): $11 + 4(2) = 19 \neq 27$
• B ($d=3$): $11 + 4(3) = 23 \neq 27$
• D ($d=5$): $11 + 4(5) = 31 \neq 27$

Takeaway: $T_m - T_n = (m-n)d$. Get the position gap by subtracting, never by counting terms inclusively — 3→4→5→6→7 is four jumps, not five.

Answer: B

Explanation

The fifth term sits four steps above the first ($T_5 = a + 4d$), so walk those four steps backwards from 23:

$$T_5 = a + 4d = a + 16 = 23 \implies a = 7$$

Or step down explicitly: $23 \to 19 \to 15 \to 11 \to 7$ — four subtractions of 4.

Why the others are wrong:

• D (3) subtracts five steps ($23 - 5 \times 4$) — the "$n$ instead of $n-1$" slip running in reverse.
• A (5) and C (9) both fail the forward check instantly: $5 + 16 = 21 \neq 23$ and $9 + 16 = 25 \neq 23$. They bracket the true answer to punish students who "walk back" hastily and land a step or two off.

Takeaway: $a = T_n - (n-1)d$. Whether you use the formula or literally count backwards, the number of steps between $T_5$ and $T_1$ is four — and a forward check ($7 + 16 = 23$ ✓) certifies the landing.

Answer: A

Explanation

"Insert 3 means between 2 and 18" means build a five-term AP from 2 to 18 and report the three middle terms. Three insertions make four gaps:

$$d = \frac{18-2}{4} = 4$$

The run is $2,\ 6,\ 10,\ 14,\ 18$, so the means are 6, 10, 14. Do not report the endpoints.

Why the others are wrong: write each run out and check the gaps are equal.

• C (5,10,15): $2,5,10,15,18$ has gaps $3,5,5,3$ ✗
• D (7,11,15): spacing of 4 inside, but anchored wrong — gaps $5,4,4,3$ ✗
• B (4,8,16): doubling, a geometric instinct. Gaps $2,4,8,2$ ✗

Takeaway: Inserting $k$ means makes $k+2$ terms and $k+1$ gaps, so $d = \frac{\text{last}-\text{first}}{k+1}$. Write the run out — unequal spacing is instantly visible.

Answer: D

Explanation

Strip the story: it starts at 20 and grows by 3 each row. An AP with $a=20$, $d=3$.

$$T_{15} = 20 + 14(3) = 62$$

Why 14 and not 15? Row 1 already has its 20 seats with no increase. The "+3" happens on each of the 14 steps from row 1 to row 15.

Why the others are wrong:

• C (65) is $20 + 15(3)$ — charging the increase to row 1 as well. The most common word-problem AP error.
• B (60) and A (55) are arithmetic slips: $14 \times 3 = 42$, not 40 or 35.

Takeaway: Starting value is $a$, per-step change is $d$, and the step count is (row number − 1). Translate the story into $T_n = a+(n-1)d$ before touching numbers.

Answer: C

Explanation

$a = 5$, $d = 3$, and a large $n$ — exactly the situation the formula exists for (nobody lists forty terms):

$$T_{40} = 5 + 39(3) = 5 + 117 = 122$$

Why the others are wrong: the three classic fumbles, all present and accounted for —

• B (117) is $39 \times 3$ alone — the first term $a = 5$ was never added back.
• D (125) is $5 + 40 \times 3$ — forty steps instead of thirty-nine ($n$ instead of $n-1$).
• A (120) treats the sequence as pure multiples of 3 ($3 \times 40$) — plausible-looking here only because $a$ and $d$ are small and close.

Takeaway: Big-$n$ term questions are two operations: $(n-1) \times d$, then $+a$. Each wrong option drops or distorts exactly one of the two — do both deliberately.

Section 8.3 — Geometric Sequences

In a geometric sequence every term is obtained by multiplying the previous term by a fixed common ratio $r$. The $n$th term is $T_n = ar^{n-1}$. If $|r| > 1$, the terms grow in magnitude. If $|r| < 1$, they shrink towards zero. If $r < 0$, the terms alternate in sign.

Answer: B

Explanation

In a geometric sequence the question is never "what is added?" but "what is multiplied?". Divide any term by the one before it:

$$r = \frac{T_2}{T_1} = \frac{12}{3} = 4$$

Confirm: $48/12 = 4$ and $192/48 = 4$ ✓ — the ratio is constant, which is what "geometric" means.

Why the others are wrong:

• C (9) is $12 - 3$ — subtracting instead of dividing, the arithmetic-sequence reflex applied to a geometric one.
• A (3) just echoes the first term — a surface grab requiring no calculation.
• D (6) is $\sqrt{3 \times 12}$, the geometric mean of the first two terms — a related idea, but the mean of two terms is not the ratio between them.

Takeaway: $r = T_{n+1}/T_n$, verified on a second pair. The first question to ask of any sequence: is it add-the-same (AP) or multiply-by-the-same (GP)? Every later formula depends on getting that classification right.

Answer: A

Explanation

Same journey logic as the AP, but the steps are multiplications: to reach the 5th term you start at 2 and multiply by 3 exactly four times.

$$T_5 = ar^{5-1} = 2 \times 3^4 = 2 \times 81 = 162$$

Walk it: $2 \to 6 \to 18 \to 54 \to 162$ — four hops ✓.

Why the others are wrong:

• B (486) is $2 \times 3^5$ — five multiplications instead of four: the GP version of the "$n$ vs $n-1$" slip.
• C (54) is $2 \times 3^3$ — one multiplication short ($T_4$, in fact).
• D (96) is $3 \times 2^5$ — the roles of $a$ and $r$ swapped (starting at 3 and doubling). Read carefully which number is the first term and which is the ratio.

Takeaway: $T_n = ar^{n-1}$ — the exponent counts the jumps, and there are always $n-1$ of them. When the numbers are small, walking the chain term by term is the perfect anti-slip check.

Answer: D

Explanation

$a=5$, $r=2$. Set the general term to 640 and divide out the first term:

$$5 \times 2^{n-1} = 640 \implies 2^{n-1} = 128$$

List the powers: $2, 4, 8, 16, 32, 64, 128$ — that is $2^7$. So $n-1 = 7$ and $n = 8$.

Check: $5 \times 2^7 = 640$ ✓

Why the others are wrong:

• B (7th) reported $n-1$ and forgot the final "+1".
• C (9th) added 1 to a finished answer.
• A (6th) miscounted the ladder, taking 128 as $2^5$.

Takeaway: Isolate $r^{n-1}$, write it as a power of $r$, then remember that exponent is $n-1$, not $n$. Listing the powers takes ten seconds and makes miscounting impossible.

Answer: C

Explanation

Find the ratio including its sign: $r = -12/4 = -3$. Alternating signs are the fingerprint of a negative ratio.

$$T_5 = 4 \times (-3)^4 = 4 \times 81 = 324$$

An even number of negatives multiplies to a positive. Or just extend the pattern: the signs run $+,-,+,-,+$, so the 5th term is positive.

Why the others are wrong:

• A ($-324$) right size, wrong sign — treated $(-3)^4$ as $-81$.
• D ($-108$) is just $T_4$ repeated.
• B ($216$) used a made-up ratio of $-2$. The ratio comes from dividing, not from guessing.

Takeaway: With $r<0$, decide the sign first (even power → positive, odd → negative), then the size. Two separate decisions.

Answer: B

Explanation

"Doubles every hour" is a GP with $r = 2$ — and the only real danger here is counting the doublings. After 4 hours, exactly 4 doublings have happened:

$$500 \;\to\; 1\,000 \;\to\; 2\,000 \;\to\; 4\,000 \;\to\; 8\,000$$

$$500 \times 2^4 = 500 \times 16 = 8\,000$$

(In sequence language this is $T_5$, because $T_1$ is "0 hours" — which is exactly why $a \times 2^{\text{hours}}$ is safer than $T_n$ bookkeeping here.)

Why the others are wrong:

• A (4 000) used only 3 doublings — that is 3 hours.
• C (16 000) used 5 doublings — 5 hours.
• D (2 000) multiplied by 4 — linear thinking about exponential growth.

Takeaway: After $n$ doublings, multiply by $2^n$. Count doublings, not terms — the hour-by-hour chain settles it in seconds.

Section 8.4 — Arithmetic Series

The sum of the first $n$ terms of an arithmetic sequence:

$$S_n = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(a + l)$$

where $l = T_n$ is the last term. The second form is Gauss's pairing trick: pair the first and last term, second and second-last, etc. — each pair sums to $a + l$.

Answer: A

Explanation

This is the sum of the first 10 odd numbers. With $a = 1$, $d = 2$:

$$S_{10} = \frac{10}{2}(2(1) + 9(2)) = 5(2 + 18) = 5 \times 20 = 100$$

Or use Gauss's pairing: the 10th odd number is $T_{10} = 1 + 9(2) = 19$, so $S_{10} = \frac{10}{2}(1 + 19) = 5 \times 20 = 100$.

Why the others are wrong:

• D (55) is the sum of the first 10 integers $1+2+\cdots+10$ — it ignores that this sequence skips the evens.
• B (90) is the pairing form with the last term taken as 17 ($T_9$, the 9th odd number) — an off-by-one in the final term: $5(1+17) = 90$.
• C (110) makes the opposite slip, pairing with 21 ($T_{11}$): $5(1+21) = 110$.

Takeaway: The sum of the first $n$ odd numbers is always $n^2$ — here $10^2 = 100$. It is one of the most re-used facts in the NBT's sequence questions (see Q36), so own it outright.

Answer: D

Explanation

The sum of the first $n$ positive integers has the famous closed form $S_n = \dfrac{n(n+1)}{2}$ (Gauss's formula — pair 1 with $n$, 2 with $n-1$, and so on).

$$\frac{n(n+1)}{2} = 55 \implies n(n+1) = 110$$

Two consecutive integers multiplying to 110: $10 \times 11$ ✓. So $n = 10$.

Why the others are wrong: all three are neighbours on the triangular-number ladder —

• A ($n=9$): $\frac{9 \times 10}{2} = 45 \neq 55$.
• B ($n=8$): $\frac{8 \times 9}{2} = 36 \neq 55$.
• C ($n=11$): $\frac{11 \times 12}{2} = 66 \neq 55$.

If 55 isn't instantly recognisable as the 10th triangular number, the ten-second test above (does $n(n+1)$ hit 110?) separates the candidates without any guessing.

Takeaway: $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$. Memorise the small triangular numbers — 15, 55, 210 for $n = 5, 10, 20$ — the NBT re-uses them constantly.

Answer: C

Explanation

$a=4$, $d=5$, $n=8$:

$$S_8 = \frac{8}{2}\bigl(8 + 35\bigr) = 4(43) = 172$$

Cross-check with the pairing form. Last term $T_8 = 4+7(5) = 39$, so $S_8 = 4(4+39) = 172$ ✓

Why the others are wrong: A (160), B (164) and D (176) all come from the bracket $8+35 = 43$ becoming 40, 41 or 44 — slips like using $8\times5$ instead of $7\times5$. None survives the two-route check.

Takeaway: Work out $2a$ and $(n-1)d$ as two separate products before adding, then confirm with $\frac{n}{2}(a+l)$. Two routes agreeing is the strongest check you have without a calculator.

Answer: B

Explanation

With the first and last terms both known, use the pairing form — you never need $d$:

$$S_{16} = \frac{16}{2}(2 + 62) = 8 \times 64 = 512$$

The picture: pair $T_1$ with $T_{16}$, $T_2$ with $T_{15}$, and so on — eight pairs, each summing to 64.

Why the others are wrong — each has the wrong bracket:

• C (496) is $8 \times 62$ — the first term never joined its pair.
• A (480) is $8 \times 60$ — the bracket taken as a difference. Pairs are added.
• D (528) is $8 \times 66$ — the first term counted twice.

Takeaway: $S_n = \frac{n}{2}(a+l)$. Say the bracket aloud — "first plus last" — and the wrong-bracket options lose their pull.

Answer: A

Explanation

A shrinking stack is still an AP, with a negative step: $a=30$, $d=-2$.

Step 1 — Bottom row. Nine steps down: $$T_{10} = 30 + 9(-2) = 12 \text{ bricks}$$

Step 2 — Pair top with bottom. $$S_{10} = \frac{10}{2}(30+12) = 210$$

Why the others are wrong — each used a wrong last term:

• B (200) took ten steps instead of nine, giving a bottom row of 10.
• C (220) miscounted the other way, giving 14.
• D (195) fits no clean method — the sort of answer you get juggling both steps mentally.

Takeaway: Find the last term first, on its own line, then pair. This problem is won or lost on the bottom row.

Section 8.5 — Geometric Series

The sum of the first $n$ terms of a geometric sequence:

$$S_n = \frac{a(r^n - 1)}{r - 1} \quad (r \neq 1)$$

Use this form when $r > 1$. When $r < 1$, the equivalent form $\dfrac{a(1-r^n)}{1-r}$ avoids negative signs.

Answer: D

Explanation

$a = 1$, $r = 2$, six terms:

$$S_6 = \frac{1(2^6 - 1)}{2 - 1} = \frac{64 - 1}{1} = 63$$

Sanity check by brute force — the terms are small enough: $1 + 2 + 4 + 8 + 16 + 32 = 63$ ✓. Notice the pattern: each doubling sum lands one short of the next power of 2.

Why the others are wrong:

• A (32) is $2^5$ — the sixth term, not the sum of six terms. ($T_n$ vs $S_n$ confusion is error #2 on this chapter's opening list.)
• B (64) is $2^6$ — the "one short" fact forgotten: the sum is $2^6 - 1$, not $2^6$ itself.
• C (127) is $2^7 - 1 = S_7$ — one term too many.

Takeaway: For the doubling GP starting at 1, $S_n = 2^n - 1$ ("all the previous powers sum to one less than the next power"). And always ask first: is the question requesting a term or a sum?

Answer: C

Explanation

$$S_5 = \frac{2(3^5 - 1)}{3 - 1} = \frac{2 \times 242}{2} = 242$$

($3^5 = 243$: walk the powers $3, 9, 27, 81, 243$.) Brute-force check: the five terms are $2 + 6 + 18 + 54 + 162 = 242$ ✓.

Why the others are wrong:

• B (486) is $2 \times 3^5$ — a term ($T_6$), not a sum. The formula's $r^n$ lives inside $\frac{a(r^n-1)}{r-1}$; on its own it computes terms.
• A (240) is the sum with the first term missing: $6 + 18 + 54 + 162 = 240$. Every term got multiplied by $r$ once too often — the series must start at $a$, not at $ar$.
• D (244) is $242 + 2$ — the first term counted twice, the mirror image of A's error.

Takeaway: $S_5 = \frac{a(r^5-1)}{r-1}$, and with numbers this small, adding the five terms directly is a fast, decisive check. Options that differ from the truth by exactly $\pm a$ are telling you someone mislaid the first term.

Answer: B

Explanation

For the doubling GP starting at 1, the sum has the clean closed form $S_n = 2^n - 1$ (Q21). Set it equal to 127:

$$2^n - 1 = 127 \implies 2^n = 128$$

List the powers: $2, 4, 8, 16, 32, 64, 128$ — seven of them. So $n = 7$.

Why the others are wrong: the wrong options are neighbouring rungs of the $2^n - 1$ ladder —

• A ($n=6$): $S_6 = 63$, not 127.
• D ($n=5$): $S_5 = 31$ — far short.
• C ($n=8$): $S_8 = 255$ — overshoot.

As with Q8's interest ladder, these questions test recognition: if $63, 127, 255$ ring bells as "one less than a power of 2", no wrong option can touch you.

Takeaway: Add 1 to the target sum and ask "which power of $r$ is this?" — then that exponent is $n$ itself (because the series started at $a = 1$). The values $2^7 = 128$ and $S_7 = 127$ are worth owning outright.

Answer: A

Explanation

Chapter 7's annuity seen through sequence eyes. Each deposit grows for a different number of years — 2, then 1, then 0:

$$100(1.1)^2 + 100(1.1) + 100 = 100(1.21+1.10+1.00) = \text{R}331$$

The amounts $100, 110, 121$ form a GP with $r=1.1$ — an annuity's future value is a geometric series.

Why the others are wrong:

• C (R300) ignores interest entirely.
• B (R330) applies 10 % once to the whole R300, as if every deposit arrived together.
• D (R333) is careless addition — the bracket is exactly 3.31.

Takeaway: Equal end-of-period deposits form a GP with ratio $(1+i)$, and the last deposit earns nothing. Write one line per payment and the series appears by itself.

Answer: C

Explanation

A reverse sum problem. With $r=2$ and $n=4$, the total is always 15 times the first term, since $1+2+4+8 = 15$:

$$15a = 120 \implies a = 8$$

Check: $8+16+32+64 = 120$ ✓

Why the others are wrong:

• A (4) divided by 30 — summing from $ar$ instead of $a$ ($2+4+8+16$).
• D (16) divided by 7.5 — the arithmetic habit of halving, sneaking into a geometric formula.
• B (12) fails the check: $12+24+48+96 = 180$.

Takeaway: $a = \frac{S_n(r-1)}{r^n-1}$. Reverse problems give you a free check — rebuild the sequence from your $a$ and add it up.

Section 8.6 — Sigma Notation

Sigma notation $\displaystyle\sum_{k=m}^{n} f(k)$ is a compact way to write a series. The variable $k$ is the index, $m$ is the lower limit, and $n$ is the upper limit. Expand the notation by substituting $k = m, m+1, \ldots, n$ and summing.

Answer: D

Explanation

Sigma notation is an instruction, not a mystery: "substitute $k = 1$ through $5$ into the expression after the $\Sigma$, and add the results."

$$\sum_{k=1}^{5} k = 1 + 2 + 3 + 4 + 5 = 15$$

Why the others are wrong:

• A (10) stops at $k = 4$ — the upper limit is inclusive; $k = 5$ belongs in the sum.
• C (20) is $5 \times 4$, a corrupted use of the formula ($n(n-1)$ instead of $n(n+1)/2$).
• B (12) matches no correct reading — with only five tiny terms, expanding and adding leaves no room for such an option to feel plausible.

Takeaway: When limits are small, expand — always. For big limits, $\displaystyle\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$; for $n = 5$: $\frac{5 \times 6}{2} = 15$ ✓ (the formula and the expansion must agree).

Answer: B

Explanation

Substitute each $k$ and add:

$$k=1:\ 3 \qquad k=2:\ 5 \qquad k=3:\ 7 \qquad k=4:\ 9$$ $$\text{Sum} = 3 + 5 + 7 + 9 = 24$$

Formula check: this is an AP with $a = 3$, $l = 9$, $n = 4$: $S_4 = \frac{4}{2}(3 + 9) = 24$ ✓.

Why the others are wrong:

• A (20) sums $2k$ without the $+1$'s: $2 + 4 + 6 + 8 = 20$. Four dropped $+1$'s are exactly the four marks between A and the right answer.
• D (28) reads $2k + 1$ as $2(k+1)$: $4 + 6 + 8 + 10 = 28$. Brackets change everything.
• C (22) is an addition slip in the four-term sum — the pairing check $(3+9) + (5+7) = 12 + 12$ makes the true total hard to miss.

Takeaway: Substitute-and-add is the base method; the AP sum formula is the check. Watch two things like a hawk: the exact expression being summed, and every $+$constant it carries.

Answer: A

Explanation

Expand the first few terms — that is how you identify any sequence hiding in sigma notation.

$$T_1 = 3(1)-1 = 2, \qquad T_2 = 5, \qquad T_3 = 8$$

Differences of 3 and 3 — constant. So it is an AP with $a=2$, $d=3$.

Why the others are wrong:

• D took the "$-1$" as the first term. The first term comes from substituting $k=1$, giving 2.
• C swapped the roles — the 3 is the coefficient of $k$, so it is $d$, not $a$.
• B calls it a GP. A $k$ multiplied by a constant means "add that much each step"; only $k$ in an exponent multiplies.

Takeaway: Linear in $k$ → AP (difference = the coefficient). Exponential in $k$ → GP (ratio = the base). In doubt, expand three terms.

Answer: C

Explanation

The $k$ is in the exponent, so this is geometric. Expand:

$$2 + 6 + 18 + 54 = 80$$

Formula check ($a=2$, $r=3$, $n=4$): $S_4 = \frac{2(81-1)}{2} = 80$ ✓

Why the others are wrong:

• A (81) is just $3^4$ — the factor 2 and the summing both ignored.
• B (78) dropped the $k=1$ term. The exponent $k-1$ makes it $2\cdot3^0 = 2$, and $3^0=1$ catches people.
• D (84) is an addition slip; expanding and formula-checking leaves it no room.

Takeaway: $k$ in the exponent → GP. Watch the $k=1$ term (anything to the power 0 is 1), expand when $n$ is small, and let the formula confirm.

Answer: D

Explanation

The lower limit is 3, not 1 — the most-missed detail in sigma questions. Substitute $k=3$ to $7$ (five terms):

$$6 + 8 + 10 + 12 + 14 = 50$$

Structure check: five terms with middle term 10, so $5 \times 10 = 50$ ✓

Why the others are wrong:

• A (40) lost the middle term — pairing $6+14$ and $8+12$ leaves the unpaired 10 behind.
• C (42) misread both limits, summing $k=1$ to 6.
• B (45) misread the upper limit as 6.

Takeaway: Read both limits first, count the terms as $(\text{top}-\text{bottom})+1$, then expand. With an odd number of AP terms, "middle × count" is a fast check.

Section 8.7 — Infinite Geometric Series

When $|r| < 1$, the terms of a GP shrink towards zero, so the infinite sum converges to a finite value:

$$S_\infty = \frac{a}{1-r} \quad \text{only when } |r| < 1$$

If $|r| \geq 1$, the sum diverges (grows without bound) and $S_\infty$ does not exist.

Answer: B

Explanation

An infinite sum only settles at a finite value if the terms shrink towards zero. That happens exactly when $|r| < 1$, meaning $-1 < r < 1$.

Why the others are wrong — each admits or excludes the wrong ratios:

• C ($r<1$) allows $r=-5$: the terms $a, -5a, 25a, \ldots$ explode. The absolute value is the condition, not decoration.
• A ($r>0$) allows $r=2$, where the terms double forever.
• D keeps only the negative half, wrongly excluding $r=\frac12$ — the commonest convergent case of all.

Takeaway: $S_\infty$ exists only when $|r|<1$. Test any proposed condition by hunting for a ratio it wrongly lets in or leaves out.

Answer: C

Explanation

$a=12$, $r=\frac12$, and $|r|<1$ so the sum exists:

$$S_\infty = \frac{12}{1-\tfrac12} = 24$$

Watch it converge: $12, 18, 21, 22.5, 23.25, \ldots$ — halving the remaining gap each time, leaning on 24.

Why the others are wrong:

• A (12) is only the first term.
• B (18) stops after two terms; the infinite tail adds another 6.
• D (36) mangles the denominator — and the running total never passes 24, so it is impossible.

Takeaway: $S_\infty = \frac{a}{1-r}$, and dividing by a fraction means multiplying by its reciprocal. Adding three or four terms tells you roughly where the answer must land — a useful sanity rail.

Answer: A

Explanation

Reverse problem: the sum is known, the first term isn't. Rearranging $S_\infty = a/(1-r)$ gives $a = S_\infty(1-r)$:

$$a = 20\left(1 - \frac{1}{4}\right) = 20 \times \frac{3}{4} = 15$$

Check forwards: $S_\infty = \dfrac{15}{1 - \tfrac14} = \dfrac{15}{\tfrac34} = 20$ ✓. (The series $15 + 3.75 + 0.9375 + \cdots$ indeed leans on 20.)

Why the others are wrong:

• B (80) divides where it should multiply: $20 \div \tfrac14 = 80$ — running the formula in the wrong direction. Its forward check explodes: $80/(3/4) \approx 107$, not 20.
• C (5) multiplies by $r$ instead of by $(1-r)$: $20 \times \tfrac14 = 5$.
• D (10) multiplies by $\tfrac12$ — a "take half" reflex with no basis in the given $r = \tfrac14$.

Takeaway: $a = S_\infty(1 - r)$: the first term is always smaller than the infinite sum (the tail contributes the rest) — so any candidate bigger than 20, like option B, is disqualified before you compute anything.

Answer: D

Explanation

A recurring decimal is an infinite geometric series hiding in plain sight — each repeat is the previous one shifted a decimal place:

$$0.333\ldots = 0.3 + 0.03 + 0.003 + \cdots$$

$a = 0.3$, $r = 0.1$ (each term is a tenth of the one before), and $|r| < 1$:

$$S_\infty = \frac{0.3}{1 - 0.1} = \frac{0.3}{0.9} = \frac{3}{9} = \frac{1}{3}$$

Why the others are wrong:

• A ($\tfrac{3}{10}$) is $0.3$ exactly — the first term alone. Truncating a recurring decimal always undershoots.
• C ($\tfrac{3}{11}$) is what the formula produces if you add $r$ instead of subtracting: $0.3/1.1 = \tfrac{3}{11} = 0.2727\ldots$ — visibly the wrong recurring decimal.
• B ($\tfrac14$) is $0.25$, a familiar-fraction guess. Divide it out: $0.25 \neq 0.333\ldots$ Candidate fractions can always be tested by division.

Takeaway: Recurring decimal → infinite GP with $r = 0.1^{(\text{repeating digits})}$ → $a/(1-r)$. And $0.\overline{3} = \tfrac13$ deserves to be instant recall — it anchors the whole family ($0.\overline{6} = \tfrac23$, $0.\overline{9} = 1$).

Answer: B

Explanation

Solve the sum formula for $r$ — note that the fraction $a/S_\infty$ gives you $1-r$, not $r$ itself:

$$24 = \frac{8}{1 - r} \implies 1 - r = \frac{8}{24} = \frac{1}{3} \implies r = \frac{2}{3}$$

Check: $|r| = \tfrac23 < 1$ ✓ (convergent), and forwards: $\dfrac{8}{1 - \tfrac23} = \dfrac{8}{\tfrac13} = 24$ ✓.

Why the others are wrong:

• A ($\tfrac13$) is the value of $1 - r$ reported as $r$ — the solution abandoned one line before the finish. This near-miss is the whole reason the question exists.
• C ($\tfrac12$) gives $S_\infty = 8/\tfrac12 = 16 \neq 24$.
• D ($\tfrac34$) gives $S_\infty = 8/\tfrac14 = 32 \neq 24$. Both fail the same five-second forward check.