Question

The sum of an infinitely decreasing GP is 3.5 and the sum of squares in terms is 14716. The sum of the cubes of the terms of the progression is​

Answer 1

Explanation:

Given The sum of an infinitely decreasing GP is 3.5 and the sum of squares in terms is 147/16. The sum of the cubes of the terms of the progression is

• Let the Geometric Progression be a, ar, ar^2, ar^3………
• So the first term is a and the common ratio is r
• So we have  
•              Sum = a / 1 – r
•                   3.5 = a/1-r
•                a / 1 – r = 7/2 (3.5 can be written as 7/2) -------------------1
• Now sum of squares of terms is a^2, a^2r^2, a^2r^4……….
•           So Sum = a^2 / 1 – r^2
•                  147 / 16 = a^2 / 1 – r^2  
•                   a^2 / 1 – r^2 = 147 / 16 ------------- 2
• Dividing equation 2 by 1 we get
•                 a^2 / 1 – r^2 x 1 – r / a = 21 / 8
•                         a / (1 – r) (1 + r) x 1 – r = 21 / 8
•                                      a / 1 + r = 21 / 8 ----------- 3
• Dividing equation 3 by 1 we get
•                              1 + r / 1 – r = 4/3
•                           3 + 3r = 4 – 4r
•                                  r = 1/7
• Now substituting r in equation 1 we get
•                       a / 1 – r = 7/2
•                      a / 1 – 1/7 = 7/2
•                          Or a = 3
• So we get
•                   a^3 / 1 – r^3 = 3^3 / 1 – (1/7)^3
•                                         = 27 / 1 – 1/343
•                                          = 343 x 3 / 38
•                                         = 1029 / 38
• So the sum of the cubes of the terms of the progression is​ 1029 / 38

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https://brainly.in/question/11445123