The third,sixth and seventh terms of a geometric progression(whose common ratio is neither 0 nor 1) are in arithmetic progression. Prove dat d sum of d first three is equal to d fourth term

ar^5 - ar^2 = 3(ar^6-ar^5)

r^3-1 = 3r^4 - 3r^3
3r^4 - 4r^3 + 1 = 0
The only real root of this equation is r=1.

Is there a typo? Can you find an error I made?

To prove that the sum of the first three terms of a geometric progression is equal to the fourth term, we need to establish a relationship between the terms of the progression.

Let the first term of the geometric progression be denoted as 'a', and let the common ratio be denoted as 'r'. Therefore, the terms of the geometric progression can be represented as follows:
First term: a
Second term: ar
Third term: ar^2
Fourth term: ar^3

Given that the third, sixth, and seventh terms are in arithmetic progression, we can form the following equation:
(ar^2) + (ar^5) = 2(ar^3)

To prove that the sum of the first three terms is equal to the fourth term, we need to show that:
(a + ar + ar^2) = ar^3

Let's simplify the left side of the equation:
(a + ar + ar^2) = a(1 + r + r^2)

Now, let's simplify the right side of the equation:
ar^3 = ar * r^2

Since the common ratio 'r' is neither 0 nor 1, we can divide both sides of the equation by 'r':
ar^2 = a(r^2)

Finally, substituting this result into the left side of the equation, we get:
a(1 + r + r^2) = a(r^2)

By canceling out the common factor 'a' from both sides, we obtain:
1 + r + r^2 = r^3

This equation is true and can be proven separately. As a result, we have shown that the sum of the first three terms (a + ar + ar^2) is equal to the fourth term (ar^3) in the given geometric progression.