I don't know if there is a ready formula for this. Let us derive one.

Let X = amount of scholarship to be deducted every year

A = amount of investment after any year

P = principal = $50,000 here.

r = rate of interest = 0.055 here.

n = number of years

In compounded annually,

A = P(1+r)^n

After year 1,

A = P(1+r)

Atfer withdrawal of X,

A = P(1+r) -X

After year 2,

A = [P(1+r) -X](1+r) = P(1+r)^2 -X(1+r)

After withdrawal of X,

A = P(1+r)^2 -X(1+r) -X

After year 3,

A = [P(1+r)^2 -X(1+r) -X](1+r)

A = P(1+r)^3 -X[(1+r)^2 +(1+r)]

After withdrawal of X,

A = P(1+r)^3 -X[(1+r)^2 +(1+r)] -X

.

.

After year 25,

A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)]

After withdrawal of X,

A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i)

And that is now equal to zero.

The [(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] can be rewritten as

[(1+r) +(1+r)^2 +(1+r)^3 +...+(1+r)^24].

It is a geometric series where

common ratio = (1+r)

a1 = (1+r) also

n = 24

So, since (1+r) = (1+0.055) = 1.055, then,

Sn = (a1)[(1 -r^n)/(1-r)]

S(24) = (1.055)[(1 -(1.055)^24) / (1 -(1.055)]

S(24) = (1.055)[(-2.61459)/(-0.055)]

S(24) = 50.15259 ----------------------***

Substituting that and the givens into (i),

A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i)

which is zero, so,

0 = (50,000)(1.055)^25 -X(50.15259) -X

0 = 190,670 -X(51.15259)

X = (190,670)/(51.15259)

X = $3727.47 -----------------answer.