Question

if the sum of infinite terms of gp is 8 and the sum of the squares of infinite tern is 4 then find the sum of cubes of the terms​

Answer 1

Given: sum of infinite terms of gp is 8,  sum of the squares of infinite term of gp is 4.

To find: the sum of cubes of the terms​?

Solution:

• Let the first term be a, and the common ratio be r.

• So sum of an infinite GP is = a/(1-r)

• Now, squares of its terms will be

              a^2, a^2r^2, a^2r^4,…..

              here first term is a^2 and common ratio is r^2.

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

              a^2/(1-r^2)

• Now we have given that  sum of infinite terms of gp is 8

              a/(1-r) = 8  .....................................(i)

• and  sum of the squares of infinite tern is 4

              a^2/(1-r^2) = 4

              a x a / (1-r)(1+r) = 4

              8 x a/1+r = 4

              a/(1+r) = 1/2

              1+r / 1-r = 2/8

              1+r / 1-r = 1/4

• By cross multiplying, we get:

              4+4r = 1-r

              3 = -5r

              r = -3/5

              putting value of r in (i), we get:

              a/(1 + 3/5) = 8

              5a / 5+3 = 8

              5a = 64

              a = 64/5

• So sum of cubes of  a^2, a^2r^2, a^2r^4,….. = a^3 / (1 - r^3)

             64/5^3 / (1-(-3/5)^3)

             262144 / 125 + 27

             262144 / 152

             1724.631

Answer:

      So, the sum  cubes of the terms​ is 1724.631.