three numbers are in harmonic progression. If the third number were decreased by 4 they would

be in arithmetic progression. If the third number were decreased by 3 they would be in geometric
progression. Find the third term of the harmonic progression.

how do i first derive the equation?

Well, well, well, it seems like those numbers are really playing games with us, don't they?

Let's call the three numbers in harmonic progression a, b, and c. Now, according to the problem, we know that:

1) a, b, and c are in harmonic progression:
That means that the reciprocal of each number is in arithmetic progression.

2) If the third number were decreased by 4, they would be in arithmetic progression:
So, if we subtract 4 from c, the reciprocal of a, b, and c would be in arithmetic progression.

3) If the third number were decreased by 3, they would be in geometric progression:
This means that if we subtract 3 from c, the ratio between the reciprocals of a, b, and c would be constant.

Now, let's have some fun with equations!

1) From the first condition, we have:
1/a, 1/b, 1/c in arithmetic progression.

2) From the second condition, if we subtract 4 from c, we have:
1/a, 1/b, 1/(c-4) in arithmetic progression.

3) From the third condition, if we subtract 3 from c, we have:
(1/a) / (1/b), (1/b) / (1/(c-3)) in geometric progression.

Now, let's simplify the third condition:
(1/a) / (1/b) = (1/b) / (1/(c-3))
b/a = 1/(c-3)

Oh, look at that! We have an expression for b in terms of a and c! How exciting! Let's keep going.

Now, if we substitute b = a/(c-3) into the first condition, we get:
1/a, (c-3)/a, 1/c in arithmetic progression.

But wait, there's more!

If a, (c-3), and c are in arithmetic progression, that means the common difference is the same between each pair of consecutive terms.

Let's calculate that common difference by subtracting (c-3)/a from 1/a:
(1/a) - ((c-3)/a) = 1/c - (c-3)/a

After simplifying, we get:
3/a = 1/c

Ah, finally, we have an equation for a in terms of c:
3 = a/c

And since we're looking for the third term of the harmonic progression, that's c!

So, the third term of the harmonic progression is... drumroll, please... c = 3!

I hope all this mathematical clowning around made you smile!

Let's suppose the three numbers in the harmonic progression are 1/a, 1/b, and 1/c, where a, b, and c are positive numbers.

According to the information given, if the third number, 1/c, were decreased by 4, the numbers would be in arithmetic progression. This means that the differences between consecutive terms would be the same.

So, we can write the arithmetic progression as:
1/a, 1/b, 1/c-4

The difference between the second and first terms is 1/b - 1/a = (a - b)/(ab).
The difference between the third and second terms is (1/c - 4) - 1/b = (b - (c-4))/(bc).

Since the differences are the same, we can set them equal:
(a - b)/(ab) = (b - (c-4))/(bc)

Cross-multiplying, we get:
c(a - b) = ab - b(c - 4)

Expanding and rearranging the terms, we get:
ca - cb = ab - bc + 4b

Simplifying further, we have:
ca - ab = bc - cb + 4b

Factoring out common terms, we get:
a(c - b) = b(c - b) + 4b

Since a, b, and c are positive numbers, we can cancel out the (c - b) term from both sides, leaving us with:
a = b + 4

Now, we know that if the third number, 1/c, were decreased by 3, the numbers would be in geometric progression. This means that the ratio between consecutive terms would be the same.

So, we can write the geometric progression as:
1/a, 1/b, 1/c-3

The ratio between the second and first terms is 1/b / 1/a = a/b.
The ratio between the third and second terms is (1/c-3) / 1/b = b/(c-3).

Since the ratios are the same, we can set them equal:
a/b = b/(c-3)

Cross-multiplying, we get:
a(c-3) = b^2

Substituting the value of a from the previous equation (a = b+4), we have:
(b+4)(c-3) = b^2

Expanding and simplifying, we get:
bc - 3b + 4c - 12 = b^2

Rearranging the terms, we get:
b^2 - bc + 3b - 4c - 12 = 0

At this point, we have two equations:
1) a = b + 4
2) b^2 - bc + 3b - 4c - 12 = 0

We can substitute the value of a in the second equation:
(b+4)^2 - (b+4)c + 3(b+4) - 4c - 12 = 0

Expanding and simplifying, we get:
b^2 + 8b + 16 - bc - 4c + 3b + 12 - 4c - 12 = 0

Rearranging and combining like terms, we get:
b^2 + (3b - bc) + (8b - 8c) + 16 - 12 = 0

Simplifying further, we have:
b^2 + b(3 - c) + 8(b - c) + 4 = 0

At this point, we need more information to solve for a specific value of b and c.

To solve this problem, let's call the three numbers in the harmonic progression as a, b, c (where b is the middle term).

We know that three numbers are in harmonic progression if their reciprocals are in arithmetic progression. So, the reciprocals of a, b, and c can be written as 1/a, 1/b, and 1/c, respectively.

According to the problem, if the third number (c) were decreased by 4, the reciprocals should form an arithmetic progression. Therefore, we have the following relationship:

(1/a), (1/b), (1/c-4)

Since the reciprocals form an arithmetic progression, we know that the common difference between terms would be the same. So, we can set up the following equation:

(1/b) - (1/a) = (1/c-4) - (1/b)

Now, let's simplify this equation:

(1/b) - (1/a) = 1/c - 1/(c-4)

To make calculations easier, let's combine fractions on the right-hand side:

(1/b) - (1/a) = (c-c+4) / (c * (c-4))

Simplifying further:

(1/b) - (1/a) = 4 / (c * (c-4))

Since we know that the reciprocals of a, b, and c are in arithmetic progression, the left-hand side can be written as:

(a-b) / (a*b) = 4 / (c * (c-4))

Now, let's move forward with the second condition given in the problem. If the third number (c) were decreased by 3, the reciprocals should form a geometric progression. So, we have the following relationship:

(1/a), (1/b), (1/c-3)

Since the reciprocals form a geometric progression, we know that the ratio between terms would be the same. So, we can set up the following equation:

(1/b) / (1/a) = (1/c-3) / (1/b)

Now, let's simplify this equation:

(a/b) = (1/c-3) / (1/b)

To make calculations easier, let's combine the fractions on the right-hand side:

(a/b) = (1/c-3) * (b/b)

Simplifying further:

(a/b) = (b / c-3)

Now, we can rewrite the equation as:

(a/b) = b / c-3

Cross-multiplying:

a * (c-3) = b^2

Now, we have two equations:

(a-b) / (a*b) = 4 / (c * (c-4)) Equation (1)
a * (c-3) = b^2 Equation (2)

By substituting the value for a*b from Equation (2) into Equation (1), we can solve for c:

(a-b) / (b^2) = 4 / (c * (c-4))

Cross multiplying and simplifying:

(b^2) * (a-b) = 4 * c * (c-4)

Expanding and rearranging:

b^2 * a - b^3 = 4c^2 - 16c

To proceed, we need more information or values to substitute or solve the equation further.

1/b - 1/a = 1/c - 1/b

b-a = (c-4)-b
b/a = (c-3)/b

I get c=12