Find: the 12th term of the G.P. `1/(a^3x^3),\ a x ,\ a^5x^5,\ dot`

Explanation

`a = 1/(a^3x^3)`

` r = (ax)/(1/(a^3x^3))`

` = a^4x^4 `

`∴ T_(12 )= 1/(a^3x^3)(a^4x^4)^(12−1)`

` = 1/(a^3x^3)(a^4x^4)^11 `

`= a^(41)x^(41)`

so,The 12 term is `a^(41) x^(41).`