An abstract visual representation of the concept of arithmetic and geometric progressions. Depict three distinct blocks, indicating three numbers, arranged linearly. The blocks are in varying shades of blue to suggest progression. Overlaying this, subtly show the mathematical symbols for addition and multiplication, hinting at the concept of sums and progression. Please remember that the image should not contain any text.

The sum of the three numbers in Arithmetic Progression is 33. If the numbers are increased by 2, 1, and 6 respectively the new numbers will be in Geometric progression. Find these numbers.

let the 3 numbers in AP be

a-d, a, and a+d

a-d + a + a-d = 33
3a = 33
a = 11

after the changes, the new numbers are:

a-d + 2, a+1, and a+d+6
or
13-d , 12, and 17+d
which are now is GP, so

12/(13-d) = (17+d)/12
144 = 221 - 4d - d^2

d^2 + 4d - 77 = 0
(d + 11)(d-7) = 0

d = -11 or d = 7

case 1: a = 11 , d = -11
the numbers are
22, 11, and 0

case 2: a = 11, d = 7
the numbers are
4, 11, 18

checking the last one:
4, 11, and 18 are in AP
if we do the changes, the new numbers are:
6, 12, and 24
is 12/6 = 24/12 ? ----> YES

check the first one in the same way

Ah, an arithmetic progression turning into a geometric progression, a real transformation. Let's unravel this mathematical mystery!

Since the sum of the three numbers in arithmetic progression is 33, we can call the middle number "x" (because it's always the hardest to come up with names for things).

Now, let's move on to making these numbers geometric. By increasing each number by 2, 1, and 6, respectively, we get a new set of numbers that form a geometric progression.

So, let's call the first number in this geometric progression "a". The second number will be "a + 2" (because we added 2), and the third number will be "a + 2 + 1 + 6" (because we added 2, 1, and 6).

Now, here comes the fun part! In a geometric progression, the ratio between each consecutive term is the same.

Let's call this common ratio "r" (for "ratio," not "arrrrrr" like a pirate).

Using this information, we can set up the following equation:
a + (a + 2) + (a + 2 + 1 + 6) = 33

Simplifying and rearranging, we find:
3a + 11 = 33

Subtracting 11 from both sides gives us:
3a = 22

Dividing both sides by 3, we find:
a = 22/3

Now that we have the value of "a," we can determine the three numbers in the geometric progression by using the common ratio "r."

The first number, "a," is 22/3.
The second number is "a + 2" which is 22/3 + 2.
The third number is "a + 2 + 1 + 6" which is 22/3 + 9.

And there you have it, the three numbers that form a geometric progression after increasing the original numbers by 2, 1, and 6, respectively. But remember, this transformation is no laughing matter, it's just math!

Let's solve this step by step.

Step 1: Let's assume the three numbers in arithmetic progression are a-d, a, and a+d, where 'a' is the middle term and 'd' is the common difference.

Step 2: The sum of these three numbers is 33, so we can write the equation as follows:
(a-d) + a + (a+d) = 33

Step 3: Simplifying the equation:
3a = 33

Step 4: Divide both sides of the equation by 3:
a = 11

Step 5: Now let's consider the new numbers in geometric progression after they are increased by 2, 1, and 6 respectively. These numbers can be represented as (11+2), (11+1), and (11+6), which simplifies to 13, 12, and 17.

Therefore, the three numbers in the arithmetic progression are 11-2, 11, and 11+2, and the corresponding numbers in the geometric progression are 13, 12, and 17.

To find the three numbers, let's first solve the equation for the sum of an arithmetic progression:

Sum of an arithmetic progression (Sn) = (n/2) * (2a + (n-1)d)

Here, 'n' represents the number of terms in the progression, 'a' represents the first term, and 'd' represents the common difference.

We know that the sum of the three numbers in the arithmetic progression is 33. Therefore:

33 = (3/2) * (2a + 2d)

Now, let's solve this equation for 'a + d':

33 = (3/2) * (2a + 2d)
33 = (3/2) * 2(a + d)
33 = 3(a + d)
(a + d) = 33/3
(a + d) = 11

Now, let's consider the new sequence formed by increasing each number by 2, 1, and 6, respectively. We need to check if these numbers form a geometric progression.

In a geometric progression, each term is obtained by multiplying the previous term by a constant called the common ratio (r).

Let the terms of the new geometric progression be (a+2), (a+d+1), (a+2d+6).

To check if these terms form a geometric progression, we need to verify if the ratio between any two consecutive terms is the same.

Let's calculate the ratios:

(a + d + 1) / (a + 2) = (a + 2d + 6) / (a + d + 1)
Cross-multiplying, we get:
(a + d + 1) * (a + d + 1) = (a + 2d + 6) * (a + 2)

Expanding both sides and simplifying, we have:
a^2 + 3ad + 2d^2 + 2a + 2d + 1 = a^2 + 4ad + 4 + 2a + 4d + 6

Simplifying further, we obtain:
3ad + 2d^2 - 4ad - 6d - 4 = 0
-d(2a + 6) + 2d^2 - 4 = 0
(2d^2 - 4d) - (2a + 6)d = 0

Now, we can solve this equation to find the values of 'a' and 'd'.