2025 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

The sequence starts at 17 and decreases by 4 each time.
This is an arithmetic sequence with first term \(a = 17\) and common difference \(d = -4\).
The formula for the \(n\)th term is:
\(17 + (n-1)(-4) = 17 - 4n + 4 = 21 - 4n\).

M1 for \(-4n + c\) or \(kn + 21\) (where \(k \neq 0\))
A1 for \(21 - 4n\) or \(-4n + 21\)

1. Calculate the cost of 5 notebooks:
\(5 \times \$1.20 = \$6.00\)

2. Calculate the cost of 3 pens:
\(3 \times \$0.85 = \$2.55\)

3. Calculate the total cost:
\(\$6.00 + \$2.55 = \$8.55\)

4. Calculate the change received from a $10 bill:
\(\$10.00 - \$8.55 = \$1.45\).

M1 for showing the total cost is $8.55
M1 for subtracting their total cost from $10.00
A1 for 1.45

Set the function equal to 11:
\(3x - 7 = 11\)
Add 7 to both sides:
\(3x = 18\)
Divide by 3:
\(x = 6\).

M1 for setting up the equation \(3x - 7 = 11\)
M1 for solving to find \(3x = 18\)
A1 for 6

Group the like terms together:
\((4a - 2a) + (-3b + 7b)\)
Simplify each group:
\(2a + 4b\).

M1 for identifying and combining like terms, showing either \(2a\) or \(4b\) correct in the final expression
A1 for \(2a + 4b\) or \(4b + 2a\)

Find the terms by substituting \(n = 1, 2, 3\):

- First term (\(n=1\)): \(1^2 - 3 = 1 - 3 = -2\)
- Second term (\(n=2\)): \(2^2 - 3 = 4 - 3 = 1\)
- Third term (\(n=3\)): \(3^2 - 3 = 9 - 3 = 6\)

The first three terms are \(-2, 1, 6\).

M1 for substituting \(n=1, 2, 3\) to find at least two terms correctly
A1 for \(-2, 1, 6\) (must be in the correct order)

To convert from euros to dollars, divide the amount in euros by 0.90:

\(540 \div 0.90 = \frac{540}{0.9} = \frac{5400}{9} = 600\)

So, €540 is equal to $600.

M1 for writing down or showing the operation \(540 \div 0.90\)
A1 for 600

Substitute \(x = 3\) into the expression:

\(g(3) = \frac{12}{3+1} = \frac{12}{4} = 3\).

M1 for substituting \(x = 3\) correctly to get \(\frac{12}{3+1}\)
A1 for 3

1. Find the highest common factor (HCF) of the numerical coefficients 12 and 8, which is 4.
2. Find the HCF of the variables \(x^2\) and \(x\), which is \(x\).
3. The common factor to pull out is \(4x\):

\(12x^2 - 8x = 4x(3x - 2)\).

M1 for identifying a partial common factor, e.g., \(4(3x^2 - 2x)\) or \(x(12x - 8)\) or \(2x(6x - 4)\)
A1 for \(4x(3x - 2)\)

Pierre receives: \(300 \times 1.15 = 345\) dollars.

M1 for \(300 \times 1.15\) or equivalent, A1 for 345.

The sequence is arithmetic with a common difference of 4. The first term is 7. Thus, the \(n\)th term is: \(7 + (n-1) \times 4 = 4n + 3\).

B1 for \(4n + c\) (where \(c \neq 3\)) or \(kn + 3\) (where \(kn \neq 4\)), B1 for \(4n + 3\).

Substitute \(x = -4\) into the function: \(f(-4) = 5 - 3(-4) = 5 + 12 = 17\).

M1 for substituting \(-4\) into the expression correctly, A1 for 17.

The highest common factor of \(12a^2b\) and \(18ab^2\) is \(6ab\). Factorising this out gives: \(6ab(2a - 3b)\).

B1 for finding a partial common factor such as \(3ab\) or \(6a\), B1 for fully factorised correct expression.

Calculate the powers first: \(5^2 = 25\), \(2^3 = 8\), \(3^2 = 9\). Then multiply and subtract: \(25 \times 8 - 9 = 200 - 9 = 191\).

M1 for evaluating at least two powers correctly, M1 for \(200 - 9\), A1 for 191.

To express \(0.00045\) in standard form, move the decimal point 4 places to the right to get \(4.5\), which gives: \(4.5 \times 10^{-4}\).

B1 for \(4.5 \times 10^k\) (where \(k \neq -4\)) or \(c \times 10^{-4}\) (where \(c \neq 4.5\)), B1 for the correct standard form.

Divide both sides by 3: \(2x - 5 = 7\). Add 5 to both sides: \(2x = 12\). Divide by 2: \(x = 6\).

M1 for expanding the bracket to \(6x - 15 = 21\) or dividing both sides by 3 to get \(2x - 5 = 7\), M1 for isolating the \(x\) term, A1 for 6.

The perimeter formula is \(2 \times (\text{length} + \text{width}) = 32\). Thus, \(\text{length} + 6 = 16\), so \(\text{length} = 10\) cm. The area is \(\text{length} \times \text{width} = 10 \times 6 = 60\text{ cm}^2\).

M1 for using the perimeter formula to find the length, A1 for length = 10, M1 for calculating area using their length, A1 for 60.

To find the midpoint of \(PQ\), calculate the average of the \(x\)-coordinates and the average of the \(y\)-coordinates. For \(x\): \(\frac{-4 + 6}{2} = 1\). For \(y\): \(\frac{7 + (-1)}{2} = 3\). Therefore, the coordinates of the midpoint are \((1, 3)\).

M1 for showing a correct method for at least one coordinate, e.g. \(\frac{-4+6}{2}\) or \(\frac{7-1}{2}\). A1 for \((1, 3)\).

First, convert the selling price from USD ($) to Euros: \(270 \div 1.25 = 270 \div \frac{5}{4} = 270 \times \frac{4}{5} = 54 \times 4 = 216\) Euros. Next, calculate the profit in Euros: \(216 - 160 = 56\) Euros. Finally, find the percentage profit: \(\frac{56}{160} \times 100 = \frac{7}{20} \times 100 = 35\%\).

M1 for converting selling price to Euros (216) or cost price to USD (200). M1 for calculating profit (56 Euros or 70 USD). A1 for 35.

Substitute \(f(x)\) into \(g(x)\): \(g(f(x)) = 2\left(\frac{3}{2x - 1}\right) + 3 = \frac{6}{2x - 1} + 3\). Set the expression equal to 5: \(\frac{6}{2x - 1} + 3 = 5 \implies \frac{6}{2x - 1} = 2\). Solve for \(x\): \(2x - 1 = 3 \implies 2x = 4 \implies x = 2\).

M1 for writing the composite function g(f(x)) as \(\frac{6}{2x-1} + 3\). M1 for setting their expression equal to 5 and solving the linear equation. A1 for 2.

Group the terms to find common factors: \((6a^2 - 15ab) - (4ac - 10bc)\). Factor out \(3a\) from the first group and \(2c\) from the second group: \(3a(2a - 5b) - 2c(2a - 5b)\). Factor out the common binomial term \((2a - 5b)\): \((3a - 2c)(2a - 5b)\).

M1 for grouping and factorising two terms correctly, e.g., \(3a(2a-5b)\) or \(-2c(2a-5b)\). A1 for \((3a - 2c)(2a - 5b)\) or equivalent.

Find the first differences: 6, 8, 10, 12. Find the second differences: 2, 2, 2. Since the second difference is constant, the sequence is quadratic with an \(n^2\) term of coefficient \(\frac{2}{2} = 1\). Subtracting \(n^2\) from each term gives: \(3 - 1^2 = 2\), \(9 - 2^2 = 5\), \(17 - 3^2 = 8\), \(27 - 4^2 = 11\). The resulting linear sequence is 2, 5, 8, 11, which has the formula \(3n - 1\). Combining these, the \(n\)th term is \(n^2 + 3n - 1\).

M1 for finding second difference of 2. M1 for subtracting \(n^2\) from terms to find the linear sequence 2, 5, 8, 11... or equivalent method. A1 for \(n^2 + 3n - 1\).

Express all terms with a base of 3: \(9^{x+1} = (3^2)^{x+1} = 3^{2x+2}\) and \(\frac{1}{27} = 3^{-3}\). Substitute these into the equation: \(3^{2x+2} \times 3^{x-2} = 3^{-3}\). Simplify using index laws: \(3^{(2x+2) + (x-2)} = 3^{-3} \implies 3^{3x} = 3^{-3}\). Equate the exponents: \(3x = -3 \implies x = -1\).

M1 for expressing 9 and 27 as powers of 3. M1 for applying index laws correctly to get \(3^{3x} = 3^{-3}\). A1 for -1.

First, rationalise the denominator of the fraction: \(\frac{4}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{4(\sqrt{5} + 1)}{5 - 1} = \frac{4(\sqrt{5} + 1)}{4} = \sqrt{5} + 1\). Now substitute this back into the original expression: \((\sqrt{5} + 1) - \sqrt{5} = 1\).

M1 for multiplying the numerator and denominator by the conjugate \(\sqrt{5}+1\). M1 for simplifying to \(\sqrt{5}+1\). A1 for 1.

In triangle ABX, the sum of angles is 180°: \(\text{Angle BAX} = 180° - 100° - 38° = 42°\). Angle BAC (which is the same as angle BAX) and angle BDC are subtended by the same arc BC. Therefore, by circle theorems, \(\text{Angle BDC} = \text{Angle BAC} = 42°\).

M1 for finding angle BAX = 180° - (100° + 38°). M1 for identifying that angle BDC = angle BAX (angles in same segment). A1 for 42.

The total number of beads is 8. The probability of picking different colours is the probability of picking Red then Blue, plus the probability of picking Blue then Red. \(P(\text{Red, Blue}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\). \(P(\text{Blue, Red}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\). Total probability = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}.

M1 for identifying the two scenarios (Red then Blue, and Blue then Red). M1 for calculating one correct product of probabilities without replacement, e.g., \(\frac{5}{8} \times \frac{3}{7}\). A1 for \(\frac{15}{28}\) or equivalent fraction.

1. Convert Euros to SGD: \(600 \times 1.45 = 870\text{ SGD}\).
2. Convert SGD back to Euros: \(870 \div 1.50 = 580\text{ Euros}\).
3. Calculate the loss: \(600 - 580 = 20\text{ Euros}\).

M1 for \(600 \times 1.45\)
M1 for their \(870 \div 1.50\)
A1 for 20

First, find \(g^{-1}(x)\):
\(y = \frac{x+1}{2} \implies 2y = x+1 \implies x = 2y-1 \implies g^{-1}(x) = 2x-1\).
Next, find \(f(g(x))\):
\(f(g(x)) = 3\left(\frac{x+1}{2}\right) - 2 = \frac{3x+3-4}{2} = \frac{3x-1}{2}\).
Set \(f(g(x)) = g^{-1}(x)\):
\(\frac{3x-1}{2} = 2x-1 \implies 3x-1 = 4x-2 \implies x = 1\).

M1 for finding \(g^{-1}(x) = 2x-1\)
M1 for setting up the equation \(\frac{3x-1}{2} = 2x-1\)
A1 for 1

Multiply both sides by \(2v-1\):
\(u(2v-1) = v+3 \implies 2uv - u = v+3\).
Rearrange to group terms with \(v\) on one side:
\(2uv - v = u+3\).
Factorise \(v\):
\(v(2u-1) = u+3\).
Divide by \(2u-1\):
\(v = \frac{u+3}{2u-1}\).

M1 for multiplying by \(2v-1\)
M1 for grouping terms in \(v\) on one side
A1 for \(\frac{u+3}{2u-1}\)

The numerators are \(3, 7, 11, 15, \dots\) which is an arithmetic sequence with \(n\)-th term \(4n - 1\). For \(n = 10\), the numerator is \(4(10) - 1 = 39\).
The denominators are \(5, 10, 17, 26, \dots\) which can be represented by \((n+1)^2 + 1\). For \(n = 10\), the denominator is \((10+1)^2 + 1 = 121 + 1 = 122\).
Thus, the 10th term is \(\frac{39}{122}\).

M1 for identifying the 10th numerator is 39
M1 for identifying the 10th denominator is 122
A1 for \(\frac{39}{122}\)

The value of the investment at the end of 2 years is:
\(4000 \times (1.03)^2 = 4000 \times 1.0609 = \$4243.60\).
She spends \(15\%\) on a computer, leaving \(85\%\) of the money:
\(4243.60 \times 0.85 = \$3607.06\).

M1 for \(4000 \times (1.03)^2\)
M1 for their \(4243.60 \times 0.85\)
A1 for 3607.06

As \(x \to \infty\), \(2^{-x} \to 0\), so \(f(x) \to b\).
Given the horizontal asymptote is \(y = 3\), we have \(b = 3\).
Since the graph passes through \((0, 8)\):
\(f(0) = a \cdot 2^0 + b = a + 3 = 8 \implies a = 5\).
Thus, the function is \(f(x) = 5 \cdot 2^{-x} + 3\).
Evaluating at \(x = -1\):
\(f(-1) = 5 \cdot 2^1 + 3 = 10 + 3 = 13\).

B1 for identifying \(b = 3\)
M1 for finding \(a = 5\)
A1 for 13

Multiply the entire equation by \((x+2)(x-1)\):
\(8(x-1) - (x+2) = (x+2)(x-1)\)
\(8x - 8 - x - 2 = x^2 + x - 2\)
\(7x - 10 = x^2 + x - 2\)
\(x^2 - 6x + 8 = 0\)
\((x-2)(x-4) = 0\)
The solutions are \(x = 2\) and \(x = 4\).
The larger value of \(x\) is 4.

M1 for eliminating the denominators
M1 for the quadratic equation \(x^2 - 6x + 8 = 0\)
A1 for 4

From the given terms, set up simultaneous equations:
For \(n = 2\): \(a(2)^2 + b(2) = 14 \implies 4a + 2b = 14 \implies 2a + b = 7\).
For \(n = 5\): \(a(5)^2 + b(5) = 65 \implies 25a + 5b = 65 \implies 5a + b = 13\).
Subtracting the first equation from the second:
\(3a = 6 \implies a = 2\).
Substituting \(a = 2\) into the first equation:
\(2(2) + b = 7 \implies b = 3\).
Thus, the general term is \(u_n = 2n^2 + 3n\).
Evaluating for the tenth term (\(n = 10\)):
\(u_{10} = 2(10)^2 + 3(10) = 200 + 30 = 230\).

M1 for setting up simultaneous equations
M1 for finding both \(a = 2\) and \(b = 3\)
A1 for 230

We use the simple interest formula for the total amount \(A\):
\(A = P(1 + rt)\)
where \(r = 0.05\) (rate of interest as a decimal) and \(t = 4\) (time in years).

Substitute the given values into the formula:
\(1440 = P(1 + 0.05 \times 4)\)
\(1440 = P(1 + 0.20)\)
\(1440 = 1.2P\)

Solve for \(P\):
\(P = \frac{1440}{1.2} = \frac{14400}{12} = 1200\)

M1 for setting up the equation \(P(1 + 0.05 \times 4) = 1440\) or equivalent.
M1 for simplifying to \(1.2P = 1440\) or \(P = \frac{1440}{1.2}\).
A1.57 for the correct final answer 1200.

To find the inverse function, let \(y = \mathrm{f}(x)\):
\(y = \frac{5x - 2}{2x + 3}\)

Multiply both sides by \((2x + 3)\) to clear the fraction:
\(y(2x + 3) = 5x - 2\)
\(2xy + 3y = 5x - 2\)

Rearrange the terms to group all \(x\) terms on one side:
\(2xy - 5x = -3y - 2\)

Factorise \(x\) from the left-hand side:
\(x(2y - 5) = -(3y + 2)\)

Divide by \((2y - 5)\):
\(x = \frac{-(3y + 2)}{2y - 5} = \frac{3y + 2}{5 - 2y} Replace \)y\) with \(x\) to get the inverse function:
\(\mathrm{f}^{-1}(x) = \frac{3x + 2}{5 - 2x}\)

M1 for setting up \(y = \frac{5x - 2}{2x + 3}\) and rearranging to clear the fraction: \(y(2x + 3) = 5x - 2\).
M1 for collecting terms in \(x\) and factorising: \(x(2y - 5) = -3y - 2\) or equivalent.
A1.57 for final answer \(\frac{3x + 2}{5 - 2x}\) (or equivalent form).

To subtract the fractions, find a common denominator, which is \((x - 2)(2x + 1)\):
\(\frac{4}{x-2} - \frac{3}{2x+1} = \frac{4(2x + 1) - 3(x - 2)}{(x - 2)(2x + 1)} Expand the numerator: \)4(2x + 1) - 3(x - 2) = 8x + 4 - 3x + 6\)

Simplify the numerator by collecting like terms:
\(8x - 3x + 4 + 6 = 5x + 10\)

Thus, the single fraction is:
\(\frac{5x + 10}{(x - 2)(2x + 1)}\)

We can also factorise the numerator:
\(\frac{5(x + 2)}{(x - 2)(2x + 1)}\)

M1 for placing over a common denominator: \(\frac{4(2x + 1) - 3(x - 2)}{(x - 2)(2x + 1)}\).
M1 for expanding the numerator correctly to obtain \(8x + 4 - 3x + 6\).
A1.57 for the correct simplified fraction \(\frac{5x + 10}{(x - 2)(2x + 1)}\) or \(\frac{5(x + 2)}{(x - 2)(2x + 1)}\).

Let us look at the differences between successive terms of the sequence:
First differences: \(8-3 = 5\), \(15-8 = 7\), \(24-15 = 9\).
Second differences: \(7-5 = 2\), \(9-7 = 2\).

Since the second differences are constant, the sequence is quadratic of the form \(an^2 + bn + c\).
The coefficient of \(n^2\) is half of the second difference:
\(a = \frac{2}{2} = 1\)
So the term contains \(n^2\).

Let us subtract \(n^2\) from each term in the sequence:
For \(n = 1\): \(3 - 1^2 = 2\)
For \(n = 2\): \(8 - 2^2 = 4\)
For \(n = 3\): \(15 - 3^2 = 6\)
For \(n = 4\): \(24 - 4^2 = 8\)

The resulting linear sequence is \(2, 4, 6, 8, \dots\), which has the \(n\)-th term \(2n\).

Therefore, the \(n\)-th term of the original sequence is:
\(n^2 + 2n\)

M1 for finding first and second differences to establish a quadratic relation with an \(n^2\) term.
M1 for establishing the linear component \(2n\).
A1.57 for the final answer \(n^2 + 2n\) or \(n(n+2)\).

Express both bases, 4 and 32, as powers of 2:
\(4 = 2^2\)
\(32 = 2^5\)

Substitute these into the equation:
\((2^2)^{2x + 1} = (2^5)^{x - 3}\)

Apply the power of a power rule \((a^m)^n = a^{mn}\):
\(2^{2(2x + 1)} = 2^{5(x - 3)}\)
\(2^{4x + 2} = 2^{5x - 15}\)

Since the bases are the same, equate the exponents:
\(4x + 2 = 5x - 15\)

Solve for \(x\):
\(2 + 15 = 5x - 4x\)
\(17 = x\)
\(x = 17\)

M1 for expressing both sides as powers of 2: \((2^2)^{2x + 1} = (2^5)^{x - 3}\).
M1 for equating the exponents: \(2(2x + 1) = 5(x - 3)\) or \(4x + 2 = 5x - 15\).
A1.57 for final answer 17 or \(x = 17\).

Elena's interest earned over 6 years is calculated using the formula: \(\text{Interest} = \frac{P \times R \times T}{100}\). Substituting the values: \(\text{Interest} = \frac{4500 \times 3.2 \times 6}{100} = 864\). The total value is the principal plus interest: \(4500 + 864 = 5364\).

M1 for interest calculation: \(4500 \times 0.032 \times 6\) (or 864) M1 for adding the interest to the principal: \(4500 + \text{their interest}\) A1 for 5364

Substitute \(x = -3\) into the function: \(f(-3) = 3(-3)^2 - 5(-3) + 2\). Simplify the terms: \(f(-3) = 3(9) + 15 + 2 = 27 + 15 + 2 = 44\).

M1 for correct substitution: \(3(-3)^2 - 5(-3) + 2\) A1 for simplifying the calculation: \(27 + 15 + 2\) A1 for 44

Expand the brackets: \((2x - 3)(x + 5) = 2x^2 + 10x - 3x - 15\). Combine like terms: \(2x^2 + 7x - 15\).

M1 for expanding to get at least three correct terms out of \(2x^2, 10x, -3x, -15\) A1 for simplifying the linear terms to \(7x\) A1 for the correct final trinomial \(2x^2 + 7x - 15\)

The difference between consecutive terms is \(11 - 4 = 7\), which indicates a linear sequence of the form \(7n + c\). Using the first term where \(n = 1\): \(7(1) + c = 4 \implies c = -3\). Therefore, the \(n\)-th term is \(7n - 3\).

M1 for finding the common difference of 7 M1 for the term of the form \(7n + k\) where \(k\) is any constant A1 for \(7n - 3\)

The reduced price can be found by calculating 85% of the original price: \(850 \times (1 - 0.15) = 850 \times 0.85 = 722.50\).

M1 for finding the reduction: \(850 \times 0.15 = 127.50\) or for using the multiplier \(0.85\) M1 for subtracting the reduction from the original price: \(850 - 127.50\) A1 for 722.50 (accept 722.5)

Substitute \(x = 2.5\) into the equation: \(y = 4^{2.5} - 3\). Since \(4^{2.5} = (4^{1/2})^5 = 2^5 = 32\), we have \(y = 32 - 3 = 29\).

M1 for correct substitution of \(x = 2.5\) into the function M1 for evaluating \(4^{2.5} = 32\) A1 for 29

Multiply both sides of the equation by 4: \(3x - 1 = 20\). Add 1 to both sides: \(3x = 21\). Divide both sides by 3: \(x = 7\).

M1 for isolating the numerator: \(3x - 1 = 20\) M1 for isolating the term with \(x\): \(3x = 21\) A1 for 7

To find the 5th term, substitute \(n = 5\) into the \(n\)-th term formula: \(3(5^2) - 2 = 3(25) - 2 = 75 - 2 = 73\).

M1 for substituting \(n = 5\) into the expression: \(3(5)^2 - 2\) A1 for evaluating the power and multiplication: \(75 - 2\) A1 for 73

Simple Interest = \( \frac{P \times R \times T}{100} = \frac{4200 \times 3.25 \times 6}{100} = 819 \). Total value = \( 4200 + 819 = 5019 \).

M1 for \( \frac{4200 \times 3.25 \times 6}{100} \), A1 for 819 (interest), A1 for 5019 (total value)

Profit = \( 108 - 75 = 33 \). Percentage profit = \( \frac{33}{75} \times 100\% = 44\% \).

M1 for finding the profit of 33, M1 for \( \frac{33}{75} \times 100 \), A1 for 44

Evaluating each function: \(f(4) = 3(4) - 5 = 7\) and \(g(3) = 2(3^2) + 1 = 19\). Thus, \(f(4) + g(3) = 7 + 19 = 26\).

M1 for evaluating \(f(4) = 7\), M1 for evaluating \(g(3) = 19\), A1 for 26

Set the function equal to 18: \(5x - 12 = 18 \Rightarrow 5x = 30 \Rightarrow x = 6\).

M1 for setting up \(5x - 12 = 18\), M1 for rearranging to \(5x = 30\), A1 for 6

Expanding the brackets gives \(6x - 15 - 2x + 8\). Combining like terms yields \(4x - 7\).

M1 for \(6x - 15\), M1 for \(-2x + 8\), A1 for \(4x - 7\)

The highest common factor of 12 and 18 is 6. The highest common factor of \(x^2y\) and \(xy^2\) is \(xy\). Factoring out \(6xy\) gives \(6xy(2x - 3y)\).

M1 for identifying a common factor of at least \(6\), \(x\), or \(y\), A1 for the correct terms inside bracket \((2x - 3y)\), A1 for the fully factorised expression \(6xy(2x - 3y)\)

The sequence has a constant difference of 7. The general term is of the form \(7n + c\). Since the first term is 8, \(7(1) + c = 8 \Rightarrow c = 1\). Therefore, the nth term is \(7n + 1\).

M1 for finding the common difference of 7, M1 for setting up \(7n + c\), A1 for \(7n + 1\)

Term 1 = 5. Term 2 = \(5 \times 2 - 3 = 7\). Term 3 = \(7 \times 2 - 3 = 11\). Term 4 = \(11 \times 2 - 3 = 19\).

M1 for finding Term 2 = 7, M1 for finding Term 3 = 11, A1 for Term 4 = 19

To find the total value of the investment, \(A\), after 4 years, use the compound interest formula: \(A = P\left(1 + \frac{r}{100}\right)^t\). Substituting the given values: \(A = 6500(1.023)^4 \approx 7118.949169\). Subtracting the principal to find the interest earned: \(\text{Interest} = 7118.949169 - 6500 = 618.949169\). Rounding to the nearest cent gives \(618.95\).

M1 for \(6500 \times (1.023)^4\) oe. A1 for \(7118.95\) (or \(7118.949...\)). A1 for \(618.95\).

To find the first three terms, substitute \(n = 1\), \(n = 2\), and \(n = 3\) into the expression \(2n^2 - 5\): For \(n = 1\): \(2(1)^2 - 5 = -3\). For \(n = 2\): \(2(2)^2 - 5 = 3\). For \(n = 3\): \(2(3)^2 - 5 = 13\). The first three terms are \(-3, 3, 13\).

B1 for each correct term: \(-3\), \(3\), and \(13\).