Question

Find four numbers in G.P. such that sum of
the middle two numbers is 10/3 and their
product is 1.​

Answer 1

Answer:

27  ,  3   ,  1/3  , 1/27

1/27  ,  1/3   ,  3  , 27

Step-by-step explanation:

Let say 4 numbers are

a , ar , ar² , ar³

Product of middle two = 1

=> ar * ar² = 1

=> a²r³ = 1

=> (ar)² = 1/r

sum of the middle two numbers is 10/3

=> ar + ar² = 10/3

=> ar ( 1 + r) = 10/3

Squaring both sides

=> (ar)² (1 + r)² = 100/9

=>  (1 + r)²  / r = 100/9

=> 9 ( 1 + r² + 2r) = 100r

=> 9r² - 82r + 9 = 0

=> 9r² - 81r - r + 9 = 0

=> 9r(r - 9) - 1(r - 9) = 0

=> (9r - 1)(r - 9) = 0

=> r = 1/9   or r = 9

a²r³ = 1  => a² (1/9)³ = 1  => a² = 729  => a = 27

GP is    27  ,  3   ,  1/3  , 1/27

a²r³ = 1  => a² (9)³ = 1  => a² = 1/729  => a = 1/27

GP is    1/27  ,  1/3   ,  3  , 27