(2)  $\because$ $\sum\limits_{k = 0}^{12} {{a_{4k + 1}}}  = 416 \Rightarrow \frac{{13}}{2}\left[ {2{a_1} + 48d} \right] = 416$

$ \Rightarrow {a_1} + 24d = 32\,\,\,\,\,\,\,\,……….\left( 1 \right)$

Now,  ${a_9} + {a_{43}} = 66 \Rightarrow 2{a_1} + 50d = 66\,\,\,\,…….\left( 2 \right)$

form eq. $(1)$ & $(2)$ we get; $d=1$ and ${a_1} = 8$

Also, $\sum\limits_{r = 1}^{17} {a_r^2 = \sum\limits_{r = 1}^{17} {{{\left[ {8 + \left( {r – 1} \right)1} \right]}^2} = 140\,m} } $

$ \Rightarrow \sum\limits_{r = 1}^{17} {{{\left( {r + 7} \right)}^2} = 140\,m} $

$ \Rightarrow \sum\limits_{r = 1}^{17} {\left( {{r^2} + 14r + 49} \right)}  = 140\,m$

$ \Rightarrow \left( {\frac{{17 \times 18 \times 35}}{6}} \right) + 14\left( {\frac{{17 \times 18}}{2}} \right) + \left( {49 \times 17} \right) = 140$

$ \Rightarrow m = 34$