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For geometric progression multiplication of 5 terms is 1 and 5 th term is 81 times then the 1 th term.

Solution:
Your problem `->` For geometric progression multiplication of 5 terms is 1 and 5 th term is 81 times then the 1 th term.


Let the terms are `(a / r^2) , (a / r) , a , (a × r) , (a × r^2)`

Multiplication : `(a / r^2) * (a / r) * a * (a × r) * (a × r^2) = 1`

`=> a^5 = 1`

`=> a = 1`

Now `5^(th)` term is `81` times than the `1^(st)` term

`=> (a × r^2) = 81 (a / r^2)`

`=> r^4 = 81`

`=> r = 3`

Now, `a = 1` and `r = 3 => (1 / 3^2) , (1 / 3) , 1 , (1 × 3) , (1 × 3^2) => 0.11 , 0.33 , 1 , 3 , 9`






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