Problem: If the third and fourth terms of an arithmetic sequence are increased by 3 and 8 respectively, then the first four terms form a geometric sequence. Find i) the sum of the first four terms of AP ii) the second term of the GP?

Solution:
Let us assume that the first term of the arithmetic sequence is 'a' and the common difference is 'd'. Then, the four terms of the arithmetic sequence are:

a, a + d, a + 2d, a + 3d

After increasing the third and fourth terms by 3 and 8 respectively, we get the following sequence:

a, a + d, a + 2d + 3, a + 3d + 8

It is given that the first four terms form a geometric sequence. Therefore, we can write:

(a + d) / a = (a + 2d + 3) / (a + d) = (a + 3d + 8) / (a + 2d + 3) = r (let)

From the first ratio, we get:

a + d = ar

From the second ratio, we get:

a + 2d + 3 = ar^2

From the third ratio, we get:

a + 3d + 8 = ar^3

Now, we can solve for 'a' and 'd' using the first two equations:

a + d = ar

a + 2d + 3 = ar^2

Subtracting the first equation from the second, we get:

d + 3 = ar(r - 1)

Substituting the value of 'a + d' from the first equation, we get:

d + 3 = r(a + d - d)

d + 3 = r(a)

Substituting this value in the first equation, we get:

a + r(a) = ar

a = r(a) - r(a - d)

a = rd

Therefore, the first four terms of the arithmetic sequence are:

rd, rd + d, rd + 2d, rd + 3d

We know that the first four terms form a geometric sequence. Therefore,

(rd + d) / rd = (rd + 2d) / (rd + d) = (rd + 3d) / (rd + 2d) = r

Solving, we get:

r = 1 + (sqrt(5)) / 2

Therefore, the first four terms of the arithmetic sequence are:

a = rd = d((sqrt(5)) - 1) / 2

d = (a + 3d + 8) / (r^3)

Substituting the values of 'a' and 'r', we get:

d = 2

Therefore, the first four terms of the arithmetic sequence are:

a = d((sqrt(5)) - 1) / 2 = (sqrt(5) - 1)

The four terms of the geometric sequence are:

(sqrt(5) - 1), 3sqrt(5), 9 + 3sqrt(5), 27 + 12sqrt(5)

Therefore, the sum of the first four terms of the arithmetic sequence is:

a + (a + d) + (a + 2