Answer:
[tex]T_n=(4-n)\text{log}_a(3)+\text{log}_a(2)[/tex]
Step-by-step explanation:
nth term of an A.P. is given by the explicit formula,
[tex]T_n=a+(n-1)d[/tex]
Here, 'a' = First term
n = number of term
d = common difference
For an A.P. given as,
[tex]\text{log}_a(54),\text{log}_a(18),\text{log}_a(6),....[/tex]
First term 'a' of the given A.P. = [tex]\text{log}_a(54),[/tex]
Common difference 'd' = [tex]T_2-T_1[/tex]
= [tex]\text{log}_a(18)-\text{log}_a(54)[/tex]
= [tex]\text{log}_a(\frac{18}{54})[/tex]
= [tex]\text{log}_a(\frac{1}{3})[/tex]
= [tex]-\text{log}_a(3)[/tex]
[tex]T_n=\text{log}_a(54)+(n-1)[-\text{log}_a(3)][/tex]
[tex]=\text{log}_a(54)-(n-1)\text{log}_a(3)[/tex]
[tex]=\text{log}_a(3^3\times 2)-(n-1)\text{log}_a(3)[/tex]
[tex]=\text{log}_a(3^3)+\text{log}_a(2)-(n-1)\text{log}_a(3)[/tex]
[tex]=3\text{log}_a(3)+\text{log}_a(2)-(n-1)\text{log}_a(3)[/tex]
[tex]=3\text{log}_a(3)+\text{log}_a(2)-n\text{log}_a(3)+\text{log}_a(3)[/tex]
[tex]=4\text{log}_a(3)+\text{log}_a(2)-n\text{log}_a(3)[/tex]
[tex]=(4-n)\text{log}_a(3)+\text{log}_a(2)[/tex]
