SPS SPS SM Pure 2023 June — Question 18 6 marks

Exam BoardSPS
ModuleSPS SM Pure (SPS SM Pure)
Year2023
SessionJune
Marks6
TopicArithmetic Sequences and Series
TypeLogarithmic arithmetic progression
DifficultyModerate -0.8 This question tests basic logarithm laws and arithmetic series summation in a straightforward manner. Part (a) requires recognizing that ln(p^n) = n·ln(p) and summing 1+2+...+11 = 66. Part (b) uses similar techniques with ln(8) = 3ln(2). Part (c) is a simple inequality ln(2p²) < 0 leading to p < 1/√2. All steps are routine applications of standard A-level techniques with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation involved.
Spec1.04g Sigma notation: for sums of series1.06f Laws of logarithms: addition, subtraction, power rules

Given that \(p\) is a positive constant,
  1. show that $$\sum_{n=1}^{11} \ln(p^n) = k \ln p$$ where \(k\) is a constant to be found, [2]
  2. show that $$\sum_{n=1}^{11} \ln(8p^n) = 33\ln(2p^2)$$ [2]
  3. Hence find the set of values of \(p\) for which $$\sum_{n=1}^{11} \ln(8p^n) < 0$$ giving your answer in set notation. [2]

Given that $p$ is a positive constant,

(a) show that

$$\sum_{n=1}^{11} \ln(p^n) = k \ln p$$

where $k$ is a constant to be found, [2]

(b) show that

$$\sum_{n=1}^{11} \ln(8p^n) = 33\ln(2p^2)$$ [2]

(c) Hence find the set of values of $p$ for which

$$\sum_{n=1}^{11} \ln(8p^n) < 0$$

giving your answer in set notation. [2]

\hfill \mbox{\textit{SPS SPS SM Pure 2023 Q18 [6]}}