Looks tough! [part 3]

$\mathfrak{S}=1+4x+7x^2+10x^3+\dots~~~$ $x\mathfrak{S}=~~~~~~x+4x^2+7x^3+10x^4$ Substracting we get: $(1-x)\mathfrak{S}=1+3x(\color{forestgreen}{1+x^2+x^3+x^4+\dots})$ $\color{#D61F06}{\mathcal{\text{(Infinite GP)}}}\Large\color{#302B94}{(*)}$ $\large \Rightarrow \mathfrak{S}=1+\frac{3x}{\color{forestgreen}{(1-x)}}=\dfrac{2x+1}{1-x}$ Placing $\mathfrak{S}=\frac{35}{16}$ and rearranging we get : $\Large 35x^2-102x+19=0$ Using Quadratic Formula we get $x=\frac{1}{5}$ (Other value of x is rejected since it was greater than one!!) $\huge \therefore~1+5=\boxed{\color{#007fff}{6}}$

$\boxed{\color{#302B94}{*\mathfrak{NOTE:-}}\\ \text{For }\mathfrak{S} \text{ to converge |x|<1 that's why}\\\text{ series written in Green }\\ \text{is an infinite GP with common difference x< 1}}$

Using the formula from wiki notes,
$\dfrac 1 {1-r} + \dfrac {d*r}{(1 - r)^2} = \dfrac 1 {1-x} + \dfrac {3*x}{(1 - x)^2}=\dfrac {35}{16} , ~ |r|<1.\\ Solving ~ the ~ resulting ~ quadric ~ equation, 35x^2-102x+19=0, ~ we ~ get ~ x=\dfrac 1 5.\\a+b=6$