$100h + 10t + u$ ← the number
$h + t + u$ ← sum of digits, the required
The digits are in AP
$t - h = u - t$
$h - 2t + u = 0$ ← Equation (1)
If you divide the number by the sum of its digits, the quotient is 26
$\dfrac{100h + 10t + u}{h + t + u} = 26$
$100h + 10t + u= 26h + 26t + 26u$
$74h - 16t - 25u = 0$ ← Equation (2)
If the digits are reversed, the resulting number is 198 more than the original number
$100u + 10t + h = (100h + 10t + u) + 198$
$-99h + 0t + 99u = 198$ ← Equation (3)
From Equations (1), (2) and (3)
h = 2, t = 3, and u = 4
Sum of digits = 9 ← answer
Another Solution
Let d = common difference of AP
h = hundreds digit
h + d = tens digit
h + 2d = units digit
$\text{The number} = 100h + 10(h + d) + (h + 2d)$
$\text{The number} = 111h + 12d$
$\text{Reversed digits} = 100(h + 2d) + 10(h + d) + h$
$\text{Reversed digits} = 111h + 210d$
$\text{Sum of digits} = h + (h + d) + (h + 2d)$
$\text{Sum of digits} = 3h + 3d$
If you divide the number by the sum of its digits, the quotient is 26
$\dfrac{111h + 12d}{3h + 3d} = 26$
$111h + 12d = 26(3h + 3d)$
$33h = 66d$
$d = 0.5h$
Hence,
$\text{The number} = 111h + 12(0.5h) = 117h$
$\text{Reversed digits} = 111h + 210(0.5h) = 216h$
$\text{Sum of digits} = 3h + 3(0.5h) = 4.5h$
If the digits are reversed, the resulting number is 198 more than the original number
$216h = 117h + 198$
$99h = 198$
$h = 2$
$\text{Sum of digits} = 4.5(2)$
$\text{Sum of digits} = 9$ ← answer
