IB Math Analysis & Approaches SL Practice Questions -Arithmetic sequences and series-Topic: SL 1.2 Paper 1

(a) Finding the Common Difference

In an arithmetic sequence, each term is obtained by adding a constant value, called the common difference (let’s call it \(d\)), to the previous term. The nth term of an arithmetic sequence can be expressed as:

\[ u_n = a + (n-1)d \]

where \(a\) is the first term, \(d\) is the common difference, and \(n\) is the term number.

We’re told:
– The second term (\(u_2\)) is 10.
– The fourth term (\(u_4\)) is 22.

Using the formula:
– For the second term (\(n = 2\)):
\[ u_2 = a + (2-1)d = a + d = 10 \quad (1) \]
– For the fourth term (\(n = 4\)):
\[ u_4 = a + (4-1)d = a + 3d = 22 \quad (2) \]

Now we have a system of two equations:
1. \( a + d = 10 \)
2. \( a + 3d = 22 \)

To find \(d\), subtract equation (1) from equation (2):
\[ (a + 3d) – (a + d) = 22 – 10 \]
\[ 2d = 12 \]
\[ d = 6 \]

So, the common difference is 6. That’s the steady “step size” between terms—let’s confirm it makes sense later when we build the sequence.

(b) Finding the General Formula for the nth Term

Now that we have \(d = 6\), we need the first term \(a\) to write the general formula. Plug \(d = 6\) into equation (1):
\[ a + 6 = 10 \]
\[ a = 4 \]

With \(a = 4\) and \(d = 6\), the nth term formula becomes:
\[ u_n = a + (n-1)d \]
\[ u_n = 4 + (n-1) \cdot 6 \]

simplifying it:
\[ u_n = 4 + 6(n-1) \]
\[ u_n = 4 + 6n – 6 \]
\[ u_n = 6n – 2 \]

So, the expression for the nth term is \( u_n = 6n – 2 \). 

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Ans: (a) valid method to find the common difference

\(d=\frac{22-10}{2}\) OR \(10+2d=22\) OR \(u_{1}+d=10\), \(u_{1}+3d=22\) OR \(u_{3}=16\)

\(d=6\)

(b) \(u_{1}=10-6(=4)\)

\(u_{n}=4+6(n-1) \) OR \(u_{n}=6n-2\)