$x_1, x_2 ,…., x_m$ ને મધ્યક તરીકે લો, તેથી $1, x_1, x_2 ,…x_m, 31$ એ

$(m + 2)$ પદોની સમાંતર શ્રેણી છે.

હવે, $31 = T_{m + 2} = a + (m + 1)d = 1 + (m + 1)d$

$\therefore \,\,{\text{d}} = \frac{{{\text{30}}}}{{{\text{m}} + {\text{1}}}}\,\,$ આપેલ 

$:\,\,\frac{{{{\text{x}}_{\text{7}}}}}{{{{\text{x}}_{{\text{m – 1}}}}}} = \frac{5}{9}\,\,\,\,\,$

$\therefore \,\,\frac{{{{\text{T}}_{\text{8}}}}}{{{{\text{T}}_{\text{m}}}}} = \,\,\,\frac{{{\text{a}} + {\text{7d}}}}{{{\text{a}} + {\text{(m}} – {\text{1)d}}}} = \,\,\,\frac{{\text{5}}}{{\text{9}}}$

$ \Rightarrow \,\,{\text{9a}} + {\text{63d}} = {\text{5a}} + {\text{(5m}} – {\text{5)d}}$

$ \Rightarrow \,\,{\text{4}}{\text{.1}}\, = {\text{(5m}} – {\text{68)}}\,\frac{{{\text{30}}}}{{{\text{m}} + {\text{1}}}}$

$ \Rightarrow \,\,{\text{2m}} + {\text{2}} = {\text{75m}} – {\text{1020}}\,\,$

$ \Rightarrow \,{\text{73m}} = {\text{1022}}\,\,\,\,\therefore \,\,{\text{m}} = \frac{{{\text{1022}}}}{{{\text{73}}}} = {\text{14}}$