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the sum of three consecutive term of G.P is 39/10 and their product is 1 find the term s

Question

The sum of three consecutive terms of G.P is 3910 \frac{39}{10} and their product is 1. Find the terms.

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Solution

Let's denote the three consecutive terms of the geometric progression (G.P) as a/r, a, and ar.

Given that the sum of these terms is 39/10, we can write the equation:

a/r + a + ar = 39/10.

We also know that the product of the terms is 1, so we can write the equation:

(a/r) * a * ar = 1.

Solving the second equation gives us a^3 = r, so we can substitute this into the first equation to get:

a/r + a + a*r = 39/10.

Substituting a^3 for r gives us:

a/a^3 + a + a*a^3 = 39/10.

Simplifying this gives us:

1/a^2 + 1 + a^2 = 39/10.

Multiplying through by a^2 gives us:

1 + a^2 + a^4 = 39/10 * a^2.

This is a quadratic equation in a^2, which we can solve to find the values of a. Once we have the values of a, we can substitute them back into the equation a^3 = r to find the values of r.

Finally, we can substitute the values of a and r into the expressions for the terms of the G.P to find the terms themselves.

This problem has been solved

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