(i) (3+6+9+\cdots+a_{20})

A.P. with (a=3), (d=3), (n=20)

[ S_{20} = \dfrac{20}{2}\bigl[2\cdot3 + 19\cdot3\bigr] = 10(6+57) = 630 ]

Alternatively (a_{20}=3+19\cdot3=60), (S_{20}=\dfrac{20}{2}(3+60)=630)

Answer: (630)

(ii) (\dfrac{4}{\sqrt{5}}+\sqrt{5}+\dfrac{6}{\sqrt{5}}+\cdots+a_n)

Terms: (\dfrac{4}{\sqrt{5}},\ \dfrac{5}{\sqrt{5}},\ \dfrac{6}{\sqrt{5}},\ldots)

A.P. with first term (\dfrac{4}{\sqrt{5}}), common difference (\dfrac{1}{\sqrt{5}})

[ a_n = \dfrac{4}{\sqrt{5}}+(n-1)\dfrac{1}{\sqrt{5}} = \dfrac{n+3}{\sqrt{5}} ]

[ S_n = \dfrac{n}{2}\left[\dfrac{4}{\sqrt{5}}+\dfrac{n+3}{\sqrt{5}}\right] = \dfrac{n(n+7)}{2\sqrt{5}} ]

Answer: (S_n = \dfrac{n(n+7)}{2\sqrt{5}})