Question

terms of GP is 8 and sum of the squares of infinity term is 4 then find the sum of the cubes of the terms​

Answer 1

The sum of the cubes of the terms​ is 1724.631

Step-by-step explanation:

Let the first term be a, and the common ratio be r. The sum of an infinite GP is given as:

GP = a/(1 - r)

Now, squares of its terms will be

a², a²r², a²r⁴,…..

First term = a² and Common ratio = r²

So, the sum of squares of terms of GP is:

S = a²/(1 - r²)

From question, the sum of infinite terms of GP is 8

a/(1 - r) = 8 → (Equation 1)

and  sum of the squares of infinite tern is 4

a²/(1 - r²) = 4

(a × a)/((1 - r)(1 + r)) = 4

a/(1 - r) × a/(1 + r) = 4

8 × a/(1 + r) = 4

a/(1 + r) = 1/2

(1 + r)/(1 - r) = 2/8

(1 + r)/(1 - r) = 1/4

4 + 4r = 1 - r

3 = -5r

∴ r = -3/5

On substituting the value of r in equation (1), we get,

a/(1 + 3/5) = 8

5a/(5 + 3) = 8

5a/8 = 8

5a = 64

∴ a = 64/5

So sum of cubes of a², a²r², a²r⁴,….. = a³/(1 - r³)

⇒ a³/(1 - r³) = (64/5)³/(1 - (-3/5)³)

a³/(1 - r³) = (262144)/(125 + 27)

a³/(1 - r³) = (262144)/(152)

∴ a³/(1 - r³) = 1724.631