Problem: Find four numbers, the first three in A.P. and the last three in G.P., given that the sum of the first and last term is 11 and the sum of the other two terms is 10.

Solution:

Let the first term of the A.P. be a and the common difference be d. Similarly, let the last term of the G.P. be c and the common ratio be r.

Then, we have the following equations:

a + c = 11 (Sum of first and last term)
ar^2 = c (Third term of G.P.)
(a+d) + (c/r^2) = 10 (Sum of other two terms)

We can use the first two equations to eliminate c and express it in terms of a and d:

c = 11 - a

Substituting this in the third equation, we get:

(a+d) + ((11-a)/r^2) = 10

Simplifying this equation, we get:

d = (r^2-1)(10-r^2a)/((r^2-1)a-11r^2+11)

Now, we can use this value of d to find a and r. Substituting it in the equation ar^2 = c, we get:

a(r^4-1) = 11r^2 - 11

We can solve this equation for r to get:

r = +/- sqrt((11a-1)/(a+11))

Note that we need to choose the positive value of r since the terms are in increasing order.

Once we have r, we can find a using the equation d = (r^2-1)(10-r^2a)/((r^2-1)a-11r^2+11). Then, we can find c using the equation c = 11 - a.

Finally, we can write down the four numbers:

a, a+d, ar^2, and ar^4

Answer: The four numbers are a, a+d, ar^2, and ar^4, where a, d, r are given by:

a = 1
d = 3
r = sqrt(2)

So, the four numbers are:

1, 4, 8, 16