Question

For question b, how is the answer is ㏒ₐ2 + 4(n-1)㏒ₐ3?
Please help me.

Answer 1

`T_n=(4-n)\text{log}_a(3)+\text{log}_a(2)`

Explanation:

nth term of an A.P. is given by the explicit formula,

`T_n=a+(n-1)d`

Here, 'a' = First term

n = number of term

d = common difference

For an A.P. given as,

`\text{log}_a(54),\text{log}_a(18),\text{log}_a(6),....`

First term 'a' of the given A.P. =
`\text{log}_a(54),`

Common difference 'd' =
`T_2-T_1`

=
`\text{log}_a(18)-\text{log}_a(54)`

=
`\text{log}_a((18)/(54))`

=
`\text{log}_a((1)/(3))`

=
`-\text{log}_a(3)`

`T_n=\text{log}_a(54)+(n-1)[-\text{log}_a(3)]`

`=\text{log}_a(54)-(n-1)\text{log}_a(3)`

`=\text{log}_a(3^3* 2)-(n-1)\text{log}_a(3)`

`=\text{log}_a(3^3)+\text{log}_a(2)-(n-1)\text{log}_a(3)`

`=3\text{log}_a(3)+\text{log}_a(2)-(n-1)\text{log}_a(3)`

`=3\text{log}_a(3)+\text{log}_a(2)-n\text{log}_a(3)+\text{log}_a(3)`

`=4\text{log}_a(3)+\text{log}_a(2)-n\text{log}_a(3)`

`=(4-n)\text{log}_a(3)+\text{log}_a(2)`