Question

 $\sum _{k=0}^{12}{a}_{4k+1}$ =416
$\Rightarrow \frac{13}{2}$ [2a₁ + 48d] = 416
$\Rightarrow$a₁ + 24d  = 3 2 ..... (1)
a₉ + a₄₃ = 66$\Rightarrow$ 2a₁ + 50d = 66 ..... (2)
                          2a₁ + 48d = 64 by ...... (1)
                     ________________________
                    d = 1 and a₁ = 8
$\Rightarrow$ 140m =$\sum _{r =1}^{17}{a}_r^2\sum _{r=1}^{17}[8+(r-1).1{]}^2$
$\Rightarrow$ 140m =$\sum _{r =1}^{17}$(r + 7)²
$\Rightarrow$ 140m =$\sum _{r =1}^{24}{r}^2-\sum _{r=1}^7{r}^2$
$\Rightarrow$ 140m = $\frac{24.25.49}{6}-\frac{7.8.15}{6}$
$\Rightarrow$ 140m = $\frac{7.8.5}{6}$[105 - 3]
$\Rightarrow$ 140m = 280.17 $\Rightarrow$ m = 34