An AP or GP?

Relevant wiki: Arithmetic and Geometric Progressions Problem Solving

$\text {The given series is arithmetico - geometric series.}\ \\ \text {Let} S_n = 1 +\dfrac45+\dfrac{7}{5^2}+....+\dfrac{3n-2}{5^{n-1}} \\ \dfrac{S_n}{5} =\dfrac15 +\dfrac{4}{5^2}+\dfrac{7}{5^2}+....+\dfrac{3n-5}{5^{n-1}}+\dfrac{3n-2}{5^n} \\ \text {Subtract both The above Equations} \\ \dfrac{4}{5}S_n = 1 + [\dfrac35+\dfrac{3}{5^2}+....+\dfrac{3}{5^{n-1}}]-[\dfrac{3n-2}{5^n}] \\ S_n=\dfrac{5}{4}+\dfrac{5}{4}\times\dfrac{3}{5}(\dfrac{1-\dfrac{1}{5^{n-1}}}{1-\dfrac{1}{5}}) - (\dfrac{3n-2}{5^n}\times\dfrac54)\ \\ S_n = \dfrac54+\dfrac34(\dfrac{5^{n-1}-1}{4})\dfrac{1}{5^{n-2}} - \dfrac{3n-2}{4\times5^{n-1}}\ \\ S_n=\dfrac54+\dfrac{15}{16}-\dfrac{3}{16\times5^{n-2}}-\dfrac{3n-2}{20\times5^{n-2}}\ \\ S_n=\dfrac{35}{16}-(\dfrac{12n+7}{80(5^{n-2})})\ \\ \text{As n Approaches Infinity :-}\ \\S_\infty=\dfrac{35}{16}$ $\text{This is the Second Way to Solve This Problem : }\ \\S_\infty=1+\dfrac45+\dfrac{7}{5^2}+...\infty \\ \dfrac15S_\infty=\dfrac15+\dfrac{4}{5^2}+\dfrac{7}{5^3}+...\infty \\ \text{Subtract Both the Above equations , we get} \\ \dfrac45S_\infty=1+\dfrac35+\dfrac3{5^2}+...\infty \\\dfrac45S_\infty=1+\dfrac{\dfrac35}{1-\dfrac15} \\ S_\infty=\dfrac{35}{16}$

Here is a third way to solve it: Find the generating function, taking partial sums twice:

$(1,2,0,0,0,...) \longleftrightarrow 1+2x$ $(1,3,3,3,3,...) \longleftrightarrow \frac{1+2x}{1-x}$ $(1,4,7,10,...) \longleftrightarrow \frac{1+2x}{(1-x)^2}$ At $x=\frac{1}{5}$ this comes out to be $\boxed{\frac{35}{16}}$