$100h + 10t + u$ ← 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* + 2*d* = 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*