Term rules, special and geometric sequences

(a) $21$

1. The terms go up by 4 each time: $17 + 4 = 21$.

• B1 The correct answer, $21$.

(b) Add 4

1. $9 - 5 = 4$, $13 - 9 = 4$: add 4 each time.

• B1 The correct answer, Add 4.

(c) No: the terms are 1 more than a multiple of 4, and 50 is not.

1. Every term is 1 more than a multiple of 4: $5 = 4 + 1$, $9 = 8 + 1$, and so on.
2. $50 = 48 + 2$, so 50 is not in the sequence. (The sequence goes $\ldots, 45, 49, 53$.)

• M1 Continuing the sequence past 50 (49 and 53), or noticing each term is 1 more than a multiple of 4.
• C1 "No", with 49 and 53 shown (or 50 not 1 more than a multiple of 4).

(a) $64$

1. $49 = 7^2$, so the next square number is $8^2 = 64$.

• B1 The correct answer, $64$.

(b) $125$

1. The cube numbers are $1, 8, 27, 64, 125$: $5^3 = 125$.

• B1 The correct answer, $125$.

(c) $16$

1. Triangular numbers: $1, 3, 6, 10, 15, 21, 28$. 16 is not one of them (it is a square number).

• B1 The correct answer, $16$.

(a) 5

1. Subtract any term from the next one.
2. $-6$ $-$ ($-11$) = 5, and each later increase is also 5.

• B1 Correct answer: 5

(a) $4$

1. Let the first term be a. The next terms are 7, a + 7, a + 14 and 2a + 21.
2. Let the first term be a. The next terms are 7, a + 7, a + 14 and 2a + 21.
3. Use the fifth term: 2a + 21 = 29.
4. Therefore 2a = 8 and a = 4.

• P1 Let the first term be a. The next terms are 7, a + 7, a + 14 and 2a + 21.
• A1 Correct answer: $4$

(a) $45$

1. Generate the second and third terms in order.
   $$2(2\times 3+3)+3$$
2. Apply the rule once more for the fourth term.
   $$2\times 21+3$$
3. Therefore $45$.

• M1 Generate the second and third terms in order.
• M1 Apply the rule once more for the fourth term.
• A1 Correct answer: $45$

(b) $0$

1. Reverse the rule: subtract 3, then divide by 2.
   $$\frac{3-3}{2}$$
2. Therefore $0$.

• M1 Reverse the rule: subtract 3, then divide by 2.
• A1 Correct answer: $0$

(a) $3$

1. Express the third, fourth and fifth terms using first term x.
   $$2x+3\times 7=27$$
2. Solve the equation for the first term.
   $$2x=6$$
3. Therefore $3$.

• P1 Express the third, fourth and fifth terms using first term x.
• P1 Solve the equation for the first term.
• A1 Correct answer: $3$

(b) $44$

1. Add the fourth term to the fifth term.
   $$17+27$$
2. Therefore $44$.

• M1 Add the fourth term to the fifth term.
• A1 Correct answer: $44$

(a) $x = 8$

1. Write the terms in terms of $x$.
   $$3, \quad x, \quad x + 3, \quad 2x + 3, \quad 3x + 6$$
2. Set the fifth term equal to 30.
   $$3x + 6 = 30 \Rightarrow x = 8$$
3. Check: $3, 8, 11, 19, 30$.

• P1 Writing the third term as $x + 3$ and the fourth as $2x + 3$.
• P1 Forming $3x + 6 = 30$.
• A1 The correct answer, $x = 8$.

(b) $486$

1. Each term is 3 times the one before: $2, 6, 18, 54, 162, 486$.

• M1 Identifying the multiplier 3 and continuing, or writing $2 \times 3^5$.
• A1 The correct answer, $486$.

(a) $T_n + T_{n+1} = (n + 1)^2$

1. Write both triangular numbers.
   $$T_n + T_{n+1} = \frac{n(n + 1)}{2} + \frac{(n + 1)(n + 2)}{2}$$
2. Take out the common factor $\frac{n + 1}{2}$.
   $$= \frac{(n + 1)(n + n + 2)}{2} = \frac{(n + 1)(2n + 2)}{2}$$
3. $$= (n + 1)(n + 1) = (n + 1)^2$$

• M1 Writing $T_{n+1} = \frac{(n + 1)(n + 2)}{2}$.
• M1 Adding over a common denominator or factorising out $(n + 1)$.
• C1 Reaching $(n + 1)^2$ and stating it is a square.

(b) $\frac{32}{3}$

1. Multiply by $\frac{2}{3}$ each time: $81, 54, 36, 24, 16$.
2. $16 \times \frac{2}{3} = \frac{32}{3}$, which is not a whole number.
3. Reason: $81 = 3^4$, so the factor of 3 runs out after four multiplications.

• P1 Generating at least three terms correctly (54, 36, 24).
• A1 $\frac{32}{3}$ (or $10\frac{2}{3}$).