Question

6) Find three numbers in G.P. such that their
sum is 21 and sum of their squares is 189.

Answer 1

Answer:

3 6 12

or

12 6 3

Step-by-step explanation:

Let say numbers are

a  , ar , ar²

a = first term

r = common ratio

Sum = 21

=> a + ar + ar² = 21

=> a(1 + r + r²) = 21

squaring both sides

=>  a²(1 + r + r²)² = 441

sum of their squares is 189.

=> (a)² + (ar)² + (ar²)² = 189

=> a²(1  + r² + r⁴) = 189

a²(1 + r + r²)²/ a²(1  + r² + r⁴)   = 441/189

=> (1 + r + r²)² / (1  + r² + r⁴)  = 7/3

=> 3(1 + r + r²)² = 7(1  + r² + r⁴)

=> 3( r⁴ + 2r³ + 3r² + 2r + 1) = 7(1  + r² + r⁴)

=> 4r⁴ - 6r³ -2r² -6r + 4 = 0

=> 2r⁴ - 3r³ - r² - 3r + 2 = 0

=> (r - 2)(2r³ + r² + r - 1) = 0

=> (r - 2)(2r - 1)(r² + r + 1) = 0

=> r = 2 or 1/2

a(1 + r + r²) = 21

=> a(1 + 2 + 4) = 21  => a = 3

3 , 6 , 12

or

a(1 + 1/2 + 1/4) = 21 => 7a/4 = 21 => a = 12

12 , 6 , 3