SL 1.2 Arithmetic Sequences & Series HL Paper 1 – IBDP Maths AA HL – Exam Style Questions

Detail Solution

(a) 
 Step 1: Substitute \( n = 4 \) into the sum formula.
We know that \( S_n = pn^2 – qn \). Thus, for \( n = 4 \):
$$ S_4 = p(4^2) – q(4) = 16p – 4q $$
Given \( S_4 = 40 \), we have:
$$ 16p – 4q = 40 $$

Step 2: Simplify the equation.
Dividing the entire equation by 4 gives:
$$ 4p – q = 10 …………..(1)$$

Step 3: Substitute \( n = 5 \) into the sum formula.
For \( n = 5 \):
$$ S_5 = p(5^2) – q(5) = 25p – 5q $$
Given \( S_5 = 65 \), we have:
$$ 25p – 5q = 65 $$

Step 4: Simplify the second equation.
Dividing the entire equation by 5 gives:
$$ 5p – q = 13 …………..(2) $$

Step 5: Solve the system of equations.
We have the following system:
 $$ 4p – q = 10 …….(1)$$
 $$ 5p – q = 13 …….(2)$$

Subtract equation (1) from equation (2):
$$ (5p – q) – (4p – q) = 13 – 10 $$
This simplifies to:
\( p = 3 \).

Step 6: Substitute \( p \) back into one of the equations to find \( q \).
Using equation (1):
$$  4(3) – q = 10 $$
$$  12 – q = 10 $$
Thus, $$ q = 2 $$

The values we found are:
$$ p = 3   ,  q = 2 $$

(b)

step1. Find the general term $ u_{n}$ of the arithmetic sequence.
The sum of the first $n$ terms is given by:
$$S_{n} = pn^{2} – qn = 3n^{2} – 2n $$
The $n$ $n-th term  is $u_{n}$ can be expressed as:
$$u_{n} = S_{n} – S_{n-1}$$

step2. Calculate $S_{n-1}$

 using $$S_{n-1} = p(n-1)^{2} – q(n-1) = 3(n-1)^{2} – 2(n-1)$$
Expanding this gives:
$$S_{n-1} = 3(n^{2} – 2n + 1) – 2(n – 1) = 3n^{2} – 6n + 3 – 2n + 2 = 3n^{2} – 8n + 5$$

step3. Find $u_{n}$
Now, we can find $u_{n}$
$$u_{n} = S_{n} – S_{n-1} = (3n^{2} – 2n) – (3n^{2} – 8n + 5)$$
This simplifies to:
$$u_{n} = -2n + 8n – 5 = 6n – 5$$

step4. Calculate $u_{5}$

 Now we find $$u_{5} = 6(5) – 5 = 30 – 5 = 25$$

————Markscheme—————–

(a) Using \( S_4 = 40 \) and \( S_5 = 65 \):
\( 16p – 4q = 40 \)
\( 25p – 5q = 65 \)
Solving these equations, we get \( p = 3 \) and \( q = 2 \).
(b) \( u_5 = S_5 – S_4 = 65 – 40 = 25 \)