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
Answers
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