IB DP Math AA: Topic : SL 1.8: The sum of infinite geometric sequences: IB style Questions HL Paper 1

Ans: EITHER
attempt to use a ratio from consecutive terms

\(\frac{p In x}{In x} = \frac{\frac{1}{3}In x}{p In x} OR \frac{1}{3}In x = (In x)r^{2} OR p In x = In x \left ( \frac{1}{3p} \right )\)

Note: Candidates may use \(In x^{1} + In x^{p} + Inx^{\frac{1}{3}} + …..\) and consider the powers of x in geometric sequence. 

Award M1 for \(\frac{p}{1} = \frac{\frac{1}{3}}{p}.\)

OR

r = p  and r² =  \(\frac{1}{3}\)

THEN

\(p^{2}= \frac{1}{3} OR r = \pm \frac{1}{\sqrt{3}}\)

\(p = \pm \frac{1}{\sqrt{3}}\)

Note: Award M0A0 for \(r^{2} = \frac{1}{3} or p^{2} = \frac{1}{3}\)   with no other working seen. 

Ans: EITHER

\(since, \left | p \right |= \frac{1}{\sqrt{3}} and \frac{1}{\sqrt{3}}<1\)

OR

\(since, \left | p \right |= \frac{1}{\sqrt{3}} and -1<p<1\)

THEN

⇒the geometric series converges.

Note: Accept r instead of p .
          Award RO if both values of p not considered.

Ans: \(\frac{In x}{1-\frac{1}{\sqrt{3}}} (=3 + \sqrt{3})\)

\(In x = 3 – \frac{3}{\sqrt{3}}+ \sqrt{3} – \frac{\sqrt{3}}{\sqrt{3}}\)   OR \(In x = 3 – \sqrt{3} + \sqrt{3} – 1 (\Rightarrow In x =2)\)

x = e²

Ans: METHOD 1
attempt to find a difference from consecutive terms or from u₂ correct equation 

\(p In x – In x = \frac{1}{3}In x – p In x OR \frac{1}{3}In x = In x +2 (p In x – In x)\)

Note: Candidates may use \(In x^{1} + In x^{p} + In x^{\frac{1}{3}} + ……\)    and consider the powers of x in arithmetic sequence.

Award M1A1 for   \(p – 1 = \frac{1}{3} – p.\)

\(2p In x = \frac{4}{3}In x \left ( \Rightarrow 2p = \frac{4}{3} \right )\)

\(p = \frac{2}{3}\)

METHOD 2
attempt to use arithmetic mean \(u_{2} = \frac{u_{1}+u_{3}}{2}\)

\(p In x = \frac{In x + \frac{1}{3}In x}{2}\)

\(2p In x = \frac{4}{3}In x \left ( \Rightarrow 2p = \frac{4}{3} \right )\)

\(p = \frac{2}{3}\)

Ans: \(d = -\frac{1}{3}In x\)

Ans: METHOD 1

\(S_{n} = \frac{n}{2}\left [ 2 In x +(n-1)\times \left ( -\frac{1}{3}In x \right ) \right ]\)

attempt to substitute into S_n and equate to \(In \left ( \frac{1}{x^{3}} \right )\)

\(\frac{n}{2}\left \lfloor 2In x + (n-1)\times \left ( -\frac{1}{3} In x \right ) \right \rfloor = In \left ( \frac{1}{x^{3}} \right )\)

\(In\left ( \frac{1}{x^{3}} \right ) = -In x^{3}\left ( = In x^{-3} \right )\)

= −3ln x

correct working with S_n (seen anywhere)

\(\frac{n}{2}\left \lfloor 2 In x – \frac{n}{3} In x + \frac{1}{3}In x\right \rfloor OR nIn x – \frac{n(n-1)}{6}In x OR \frac{n}{2}\left ( In x + \left ( \frac{4-n}{3} \right )In x\right )\)

correct equation without ln x

\(\frac{n}{2}\left ( \frac{7}{3} – \frac{n}{3} \right ) = -3 OR n-\frac{n(n-1)}{6} = -3\) (or equivalent)

Note: Award as above if the series \(1 + p + \frac{1}{3}+ ….\)   is considered leading to \(\frac{n}{2}\left ( \frac{7}{3} -\frac{n}{3}\right ) = -3.\)

attempt to form a quadratic = 0

n² -7n − 18 = 0

attempt to solve their quadratic

(n – 9) (n + 2) = 0 

n = 9