if the 3rd term of a GP is the square of the first term and it's fifth term is 729,find the GP .
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Let the series be a,ar,ar^2...
T3=(T1)^2
ar^3-1=(ar^1-1)^2
ar^2=(ar^0)^2
ar^2=a^2
r^2=a^2/a
r^2=a
r=_/a
Now,
T5=729
ar^5-1=729
ar^4=729
a(_/a)^4=(9)^3
a^3=9^3
So, a=9
Therefore, r=_/a
r = _/9
r = 3
So, series will be 9,18,81....
T3=(T1)^2
ar^3-1=(ar^1-1)^2
ar^2=(ar^0)^2
ar^2=a^2
r^2=a^2/a
r^2=a
r=_/a
Now,
T5=729
ar^5-1=729
ar^4=729
a(_/a)^4=(9)^3
a^3=9^3
So, a=9
Therefore, r=_/a
r = _/9
r = 3
So, series will be 9,18,81....
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