ધારો કે $d\,\,\ne \,\,0\,\,$ તેથી ઉકેલ

${{\text{a}}_{\text{1}}}\,,\,\,{{a}_{2}}\,,\,\,{{a}_{3}},\,\,………\,,\,\,{{a}_{100}}\,\,\to $ સમાંતર શ્રેણી

${{\text{a}}_{\text{1}}}\,=\,\,3\,\,\,;$ ${{S}_{p}}\,=\,\,\sum\limits_{i\,=1}^{p}{{{a}_{i}}}\,\,\,\,\,\,1\,\,\le \,\,n\,\,\le \,\,20$

$m\,\,=\,\,5n$

$\frac{{{S}_{m}}}{{{S}_{n}}}\,\,=\,\,\frac{\frac{m}{2}\,[2{{a}_{1}}\,+\,\,(m\,\,-\,\,1)\,d]}{\frac{n}{2}\,[2{{a}_{1}}\,\,+\,\,(n\,\,-\,\,1)d]}$

$\Rightarrow \,\,\,\frac{{{S}_{m}}}{{{S}_{n}}}\,\,=\,\,\frac{5[(2{{a}_{1}}\,-\,\,d)\,\,+\,\,5nd]}{[(2{{a}_{1}}\,-\,\,d)\,+\,\,nd]}$

$\frac{{{S}_{m}}}{{{S}_{n}}}$ ને $\text{n}$ થી સ્વતંત્ર થવા માટે

$\therefore \,\,\text{2}{{\text{a}}_{\text{1}}}\,-\,\,d\,\,=\,\,0\,\,\,\,\Rightarrow \,\,d\,\,=\,\,2{{a}_{1}}$

$\Rightarrow \,\,d\,\,=\,\,6\,\,\,\,\,\Rightarrow \,\,{{a}_{2}}\,=\,\,9$