Let a1, a2, a3, ....., a9 be terms of a geometric series and (r) is the common difference.

Given that the 3rd and 9th are -3, and -192

But we know that `an = a1*r^(n-1)`

`` ==> `a3= a1*r^2 = -3` ...............(1)

==> `a5= a1*r^4 = -192` .............(2)

Now we...

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Let a1, a2, a3, ....., a9 be terms of a geometric series and (r) is the common difference.

Given that the 3rd and 9th are -3, and -192

But we know that `an = a1*r^(n-1)`

`` ==> `a3= a1*r^2 = -3` ...............(1)

==> `a5= a1*r^4 = -192` .............(2)

Now we will solve for a1 and r

We will divide (1) by (2).

==> `(a1r^4)/(a1r^2) = -192/(-3) `

`==gt r^2 = 64 ==gt r= +-8`

But given that the common ratio is positive. Then, we will ignore -8 and take only r=8 as a common difference.

`==gt r = 8 `

`==gt a1r^2 = -3 ==gt a1= -3/r^2 = -3/64`

`` ==> Then, the first terms is `a1= -3/64`