IBDP Maths SL 1.2 Arithmetic Sequences & Series AA HL Paper 1- Exam Style Questions- New Syllabus

(a)
For an arithmetic sequence, the \(n\)th term is given by \(u_n=a+(n-1)d\).

Here, the first term is \(a=36\) and the fifth term is \(u_5=12\).

\(12=36+(5-1)d\)

\(12=36+4d\)

\(4d=-24\)

\(d=-6\)

The common difference is \(-6\), so each term is 6 less than the previous term.

Now find the 13th term.

\(u_{13}=36+(13-1)(-6)\)

\(u_{13}=36+12(-6)\)

\(u_{13}=36-72=-36\)

✅ Answer: \(u_{13}=-36\)

(b)
The sum of the first \(n\) terms of an arithmetic sequence is \(S_n=\frac{n}{2}\left[2a+(n-1)d\right]\).

Substitute \(a=36\), \(d=-6\) and \(S_n=0\).

\(\frac{n}{2}\left[2(36)+(n-1)(-6)\right]=0\)

Since \(n\) is a positive number of terms, the expression inside the brackets must be zero.

\(72-6(n-1)=0\)

\(72-6n+6=0\)

\(78-6n=0\)

\(6n=78\)

\(n=13\)

This result also makes sense because the first and 13th terms are \(36\) and \(-36\). The terms pair symmetrically, so each pair has a sum of zero.

✅ Answer: \(n=13\)