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.

Takeaway: $r = 1 - a/S_\infty$. Whenever a rearrangement produces "$1 - r = \ldots$", underline it — the examiner is betting you'll stop there and forget the final subtraction.

Exam-Bank Extras — Question Types Confirmed in Recent Papers

The NBT MAT reuses question types from a stable bank year after year. The three questions below are modelled directly on types repeatedly confirmed in recent papers and not yet represented in this chapter. Every answer has been independently machine-verified.

Answer: B

Explanation

The rule $2n-1$ generates the odd numbers: $1, 3, 5, \ldots, 2p-1$. So this is an AP with $a=1$, $d=2$, and $p$ terms.

By formula: $$S_p = \frac{p}{2}\bigl(2 + (p-1)2\bigr) = \frac{p}{2}(2p) = p^{2}$$

By picture: $1 = 1^2$, $1+3 = 2^2$, $1+3+5 = 3^2$, $1+3+5+7 = 4^2$ — each new odd number wraps an L around the square, completing the next one.

Why the others are wrong:

• A $p(p+1)$ is the sum of the first $p$ even numbers.
• C $2p^2-p$ equals $p(2p-1)$ — the last term times $p$, a mis-expansion.
• D $p(p-1)$ is the even-number sum one step short.

Takeaway: $\sum_{n=1}^{p}(2n-1) = p^2$ — worth knowing as a single fact. "$2n\pm1$ summed from 1" should trigger "odd numbers → perfect square".

Answer: C

Explanation

Verify the rule first — never trust a stated pattern blindly: $34 = 21+13$ ✓, $21 = 13+8$ ✓, $13 = 8+5$ ✓, $8 = 5+3$ ✓. (Fibonacci, running backwards.)

Now apply it to the terms containing the unknowns:

The term 5 has right-neighbours 3 and $w$: $\ 5 = 3+w \Rightarrow w = 2$

The term 3 has right-neighbours $w$ and $r$: $\ 3 = 2+r \Rightarrow r = 1$

$$w + r = 3$$

Why the others are wrong:

• A ($2$) is $w$ alone — the question asks for $w+r$.
• D ($4$) is $w$ counted twice; the equation for $r$ was never solved.
• B ($5$) just echoes a visible term.

Takeaway: If a sequence fails both the AP test (constant differences) and the GP test (constant ratios), try an additive rule. Once one unknown falls, the rest follow one equation at a time.

Answer: D

Explanation

Both sides are arithmetic series with difference 4 and 50 terms each — plus the lone $x$ on the right.

The insight: pair the terms across the equation.

$$199 \leftrightarrow 197, \quad 195 \leftrightarrow 193, \quad \ldots, \quad 3 \leftrightarrow 1$$

Every pair differs by 2, and there are 50 pairs. So the left side exceeds the right-side series by $2 \times 50 = 100$, and $x$ must supply exactly that:

$$x = 100$$

Formula check: left $= \frac{50}{2}(3+199) = 5050$; right series $= \frac{50}{2}(1+197) = 4950$; difference $= 100$ ✓

Why the others are wrong:

• A ($50$) is the number of pairs, not the surplus.
• C ($2$) is one pair's difference; there are fifty.
• B ($200$) used the within-series difference of 4 as the per-pair gap. The gap across the equation is 2.

Takeaway: Two long series across an equals sign? Do not compute either side — pair the terms and multiply the per-pair difference by the number of pairs.

Mixed Practice Questions — M1 to M30

Answer: A

Answer: C

Answer: B

Answer: D

Answer: A

Answer: C

Answer: B

Answer: D

Answer: A

Answer: C

Answer: B

Answer: D

Answer: A

Answer: C

Answer: B

Answer: D

Answer: A

Answer: C

Answer: B

Answer: D

Answer: A

Answer: C

Answer: B

Answer: D

Answer: A

Answer: C

Answer: B

Answer: D

Answer: A

Answer: C