Number Series Questions and Answers

Answer: B

The missing term is 1152.

The terms are multiplied alternately by 4 and 3: 8 × 4 = 32, 32 × 3 = 96, 96 × 4 = 384, 384 × 3 = 1152 and 1152 × 4 = 4608, which matches the last term given.

960, 1536 and 1056 do not lead to 4608 by the alternate factors 4 and 3.

You can confirm the pattern by dividing neighbouring terms: 32 / 8 = 4, 96 / 32 = 3, 384 / 96 = 4, so the multipliers alternate between 4 and 3. The term after 384 is found by multiplying by 3, and the last step to 4608 multiplies by 4 again, which confirms 1152.

Answer: B

A possible value is 21.

The differences between consecutive terms are consecutive odd numbers: 9 - 6 = 3 and 14 - 9 = 5, so the next differences should be 7, 9 and 11. That gives \(14 + 7 = 21\), then \(21 + 9 = 30\) and \(30 + 11 = 41\), matching the terms given after x.

With 25, 18 or 20 the differences on either side of x (11 and 5, 4 and 12, 6 and 10) break the pattern, and none of them leads to 30 and 41 by a consistent rule.

Answer: A

The relations are X(i) = X(i - 1) + I(i) and Y(i) = Y(i - 1) I(i), for i > 0.

Check X: 20 + (-4) = 16, 16 + 10 = 26, 26 + 15 = 41. Each output adds the new input to the previous output.

Check Y: 20 × (-4) = -80, -80 × 10 = -800, -800 × 15 = -12000. Each output multiplies the previous output by the new input.

The option with X multiplying would give 20 × (-4) = -80 for X(1), not 16, and Y adding would give 16, not -80. The last option multiplies Y by the previous input, giving Y(1) = 20 × 20 = 400 instead of -80.

Answer: B

The missing number is 1440.

Each term is the previous one multiplied by a factor that falls by one each time: 2 × 6 = 12, 12 × 5 = 60, 60 × 4 = 240, 240 × 3 = 720, 720 × 2 = 1440. The next factor is 1, giving 1440 × 1 = 1440, and then 0, giving 1440 × 0 = 0, which matches the last term.

2880 continues doubling, which is only the last step's factor. 720 and 0 do not fit the falling multipliers, and 0 is reserved for the final term, which needs the factor 0.

Answer: A

\(F_{1}\) is 4.

The rule \(F_{n+1} = F_{n} + F_{n-1}\) can be run backwards as \(F_{n-1} = F_{n+1} - F_{n}\):

\[ \begin{aligned} F_{5} &= 60 - 37 = 23 \\ F_{4} &= 37 - 23 = 14 \\ F_{3} &= 23 - 14 = 9 \\ F_{2} &= 14 - 9 = 5 \\ F_{1} &= 9 - 5 = 4 \end{aligned} \]

Check forwards: 4, 5, 9, 14, 23, 37, 60. The value 5 is \(F_{2}\), 9 is \(F_{3}\), and 8 does not appear in the series at all.

Answer: C

The number to change is 6.

In the Fibonacci series each term is the sum of the two before it: 1, 1, 2, 3, 5, 8, 13, 21, ... Here the fifth term should be \(2 + 3 = 5\), not 6.

With 5 in place, every later term checks: \(3 + 5 = 8\), \(5 + 8 = 13\), \(8 + 13 = 21\). So 8, 13 and 21 are all correct as printed; it is only the 6 that breaks the rule.

Answer: C

The wrong term is 3962.

Each term is the square of the one before it, minus 1:

\[ \begin{aligned} 2^2 - 1 &= 3 \\ 3^2 - 1 &= 8 \\ 8^2 - 1 &= 63 \\ 63^2 - 1 &= 3968 \end{aligned} \]

The last term should be 3968, so 3962 is the one that breaks the rule. 3, 8 and 63 all fit the pattern exactly, so none of them is the wrong term. A quick check without full multiplication: \(63^2\) ends in 9, so \(63^2 - 1\) must end in 8, not 2.

Answer: C

The entry at C is 15.0.

Find the rule of the first sequence from the ratio of each term to the one before:

\[ \begin{aligned} 400/4000 &= 0.1,\ 120/400 = 0.3 \\ 60/120 &= 0.5,\ 42/60 = 0.7 \\ 37.8/42 &= 0.9 \end{aligned} \]

The multipliers are 0.1, 0.3, 0.5, 0.7, 0.9. Applying the same rule from 1000: A = 100, B = 30, C = 15, D = 10.5, E = 9.45. So 10.5 and 9.45 are the entries at D and E, and 30.0 is the entry at B; they are the off-by-one traps.

Answer: B

X is 184320.

Each term is the previous one multiplied by the next even number: \(4 \times 2 = 8\), \(8 \times 4 = 32\), \(32 \times 6 = 192\), \(192 \times 8 = 1536\), \(1536 \times 10 = 15360\).

\[ X = 15360 \times 12 = 184320 \]

The other options differ from 184320 by small amounts (−7, +1, +7) and are there to catch arithmetic slips. 184320 is also the only one divisible by 15360's factor 10, a fast check since every term after 15360 must end in 0.

Answer: C

All three rows are correctly matched.

• Row 1: 2, 6, 18, 54 multiplies by 3 each time, so the next term is \(54 \times 3 = 162\).
• Row 2: 144, 121, 100, 81 are \(12^2, 11^2, 10^2, 9^2\), so the next is \(8^2 = 64\).
• Row 3: B, D, G, K move forward 2, 3 and 4 places (B to D, D to G, G to K); the next gap of 5 takes K to P.

Each rule fits its series and each next term follows from the rule, so no row is mismatched. The trap is row 3, where a solver who counts K + 5 as O has slipped one letter.