An arithmetic sequence has the first term $\ln a$ and a common difference $\ln 3$.

The 13th term in the sequence is $8\ln 9$. Find the value of $a$.

Note:     There are many approaches to this question, and the steps may be done in any order. There are 3 relationships they may need to apply at some stage, for the 3rd, 4th and 5th marks. These are

equating bases eg recognising 9 is ${{\text{3}}^2}$

log rules: $\ln b + \ln c = \ln (bc),{\text{ }}\ln b - \ln c = \ln \left( {\frac{b}{c}} \right)$,

exponent rule: $\ln {b^n} = n\ln b$.

The exception to the FT rule applies here, so that if they demonstrate correct application of the 3 relationships, they may be awarded the A marks, even if they have made a previous error. However all applications of a relationship need to be correct. Once an error has been made, do not award A1FT for their final answer, even if it follows from their working.

Please check working and award marks in line with the markscheme.

correct substitution into ${u_{13}}$ formula     (A1)

eg$\;\;\;\ln a + (13 - 1)\ln 3$

set up equation for ${u_{13}}$ in any form (seen anywhere)     (M1)

eg$\;\;\;\ln a + 12\ln 3 = 8\ln 9$

correct application of relationships     (A1)(A1)(A1)

$a = 81$     A1     N3

[6 marks]

Examples of application of relationships

Example 1

correct application of exponent rule for logs     (A1)

eg$\;\;\;\ln a + \ln {3^{12}} = \ln {9^8}$

correct application of addition rule for logs     (A1)

eg$\;\;\;\ln (a{3^{12}}) = \ln {9^8}$

substituting for 9 or 3 in ln expression in equation     (A1)

eg$\;\;\;\ln (a{3^{12}}) = \ln {3^{16}},{\text{ }}\ln (a{9^6}) = \ln {9^8}$

Example 2

recognising $9 = {3^2}$      (A1)

eg$\;\;\;\ln a + 12\ln 3 = 8\ln {3^2},{\text{ }}\ln a + 12\ln {9^{\frac{1}{2}}} = 8\ln 9$

one correct application of exponent rule for logs relating $\ln 9$ to $\ln 3$     (A1)

eg$\;\;\;\ln a + 12\ln 3 = 16\ln 3,{\text{ }}\ln a + 6\ln 9 = 8\ln 9$

another correct application of exponent rule for logs     (A1)

eg$\;\;\;\ln a = \ln {3^4},{\text{ }}\ln a = \ln {9^2}$