TMUA Sequences & series questions

Answer: D · Method: Inclusion–exclusion: don't count twice

1. Multiples of $4$: $75$. Multiples of $6$: $50$.
2. Numbers divisible by both are the multiples of $\text{lcm}(4, 6) = 12$: there are $25$, and they were counted twice.
3. Total: $75 + 50 - 25 = 100$.

Why the tempting wrong answers are wrong

• F: You added both lists without removing the numbers counted twice (the multiples of $12$).

Answer: B · Method: Find terms from the sum formula: uₙ = Sₙ − Sₙ₋₁

1. The 17th term is the sum of the first 17 terms minus the sum of the first 16: $u_{17} = S_{17} - S_{16}$.
2. $S_{17} = 1819$ and $S_{16} = 1616$, so $u_{17} = 203$.

Why the tempting wrong answers are wrong

• D: That's $S_{18} - S_{17}$, the 18th term.
• E: That's $S_{17}$, the sum of the first 17 terms, not the 17th term.

Answer: B · Method: Find terms from the sum formula: uₙ = Sₙ − Sₙ₋₁

1. The 19th term is the sum of the first 19 terms minus the sum of the first 18: $u_{19} = S_{19} - S_{18}$.
2. $S_{19} = 437$ and $S_{18} = 396$, so $u_{19} = 41$.

Why the tempting wrong answers are wrong

• D: That's $S_{20} - S_{19}$, the 20th term.
• E: That's $S_{19}$, the sum of the first 19 terms, not the 19th term.

Answer: B · Method: Find terms from the sum formula: uₙ = Sₙ − Sₙ₋₁

1. The 13th term is the sum of the first 13 terms minus the sum of the first 12: $u_{13} = S_{13} - S_{12}$.
2. $S_{13} = 117$ and $S_{12} = 96$, so $u_{13} = 21$.

Why the tempting wrong answers are wrong

• D: That's $S_{14} - S_{13}$, the 14th term.
• E: That's $S_{13}$, the sum of the first 13 terms, not the 13th term.

Answer: B · Method: Find terms from the sum formula: uₙ = Sₙ − Sₙ₋₁

1. The 17th term is the sum of the first 17 terms minus the sum of the first 16: $u_{17} = S_{17} - S_{16}$.
2. $S_{17} = 1377$ and $S_{16} = 1216$, so $u_{17} = 161$.

Why the tempting wrong answers are wrong

• D: That's $S_{18} - S_{17}$, the 18th term.
• E: That's $S_{17}$, the sum of the first 17 terms, not the 17th term.

Answer: C · Method: Set up equations from sums to infinity

1. With first term $a$ and ratio $r$: $\dfrac{a}{1-r} = 12$ and $\dfrac{a^2}{1-r^2} = 48$.
2. Since $1 - r^2 = (1-r)(1+r)$, the second equation is $\dfrac{a}{1-r}\cdot\dfrac{a}{1+r} = 48$, so $\dfrac{a}{1+r} = 4$.
3. Then $12(1-r) = 4(1+r)$, so $r = \tfrac{1}{2}$ and $a = 6$.

Answer: C · Method: Inclusion–exclusion: don't count twice

1. Multiples of $3$: $3 + 6 + \dots + 198 = 3 \times \tfrac{66 \times 67}{2} = 6633$.
2. Multiples of $5$: $5 + 10 + \dots + 200 = 5 \times \tfrac{40 \times 41}{2} = 4100$.
3. Multiples of $15$ were counted twice: $15 \times \tfrac{13 \times 14}{2} = 1365$. Total: $6633 + 4100 - 1365 = 9368$.

Why the tempting wrong answers are wrong

• A: You subtracted the multiples of $15$ twice. They should be removed once.
• E: You didn't subtract the multiples of $15$, which are counted in both lists.

Answer: C · Method: Spot a repeating sequence

1. Work out the first few terms: $x_1 = 2$, $x_2 = -\tfrac{1}{3}$, $x_3 = \dfrac{-\frac{10}{3}}{\frac{2}{3}} = -5$, $x_4 = \dfrac{-8}{-4} = 2$. The sequence repeats every three terms.
2. Each block of three sums to $2 - \tfrac{1}{3} - 5 = -\tfrac{10}{3}$. The first $99$ terms are $33$ blocks: $33 \times \left(-\tfrac{10}{3}\right) = -110$.
3. The 100th term starts a new block, so it is $x_1 = 2$. The total is $-108$.

Why the tempting wrong answers are wrong

• B: That's the sum of the first $99$ terms. The 100th term is $2$, the start of a new cycle.

Answer: D · Method: Spot a repeating sequence

1. $x_1 = 3$, $x_2 = -\tfrac{1}{2}$, $x_3 = \tfrac{2}{3}$, $x_4 = 3$: the sequence repeats every three terms.
2. Each block sums to $3 - \tfrac{1}{2} + \tfrac{2}{3} = \tfrac{19}{6}$. The first $48$ terms are $16$ blocks: $\tfrac{304}{6}$.
3. Terms $49$ and $50$ are $3$ and $-\tfrac{1}{2}$, adding $\tfrac{15}{6}$. Total: $\tfrac{319}{6}$.

Why the tempting wrong answers are wrong

• E: That's the sum of $49$ terms. Don't forget $x_{50} = -\tfrac{1}{2}$.

Answer: C · Method: Set up equations from sums to infinity

1. With first term $a$ and ratio $r$: $\dfrac{a}{1-r} = 15$ and $\dfrac{a^2}{1-r^2} = \dfrac{225}{2}$.
2. Since $1 - r^2 = (1-r)(1+r)$, dividing the second by the first gives $\dfrac{a}{1+r} = \dfrac{15}{2}$.
3. Solving the two linear equations gives $r = \dfrac{1}{3}$ and $a = 10$.

Why the tempting wrong answers are wrong

• A: Check the sign of the ratio.

Answer: C · Method: Spot a repeating sequence

1. Work out the first few terms: $-1,\ 2,\ \dfrac{1}{2},\ -1$. The sequence repeats every 3 terms.
2. Each block of 3 sums to $\dfrac{3}{2}$. There are 16 complete blocks in the first 50 terms, giving $24$.
3. The remaining 2 terms are $-1,\ 2$, so the total is $25$.

Why the tempting wrong answers are wrong

• A: That's the sum of the first 49 terms.
• B: That's 16 complete cycles: the last 2 terms still need adding.