Question

find four number in G. P. such that sum of middle two number is 10/3 and their products is 1. ​

Answer 1

Answer:

Step-by-step explanation:

Given  

find four number in G. P. such that sum of middle two number is 10/3 and their products is 1.  

Let GP be a/r^3, a/r, ar, ar^3

So a^4 = 1

a = ± 1

Given sum of middle terms = 10/3

a/r + ar = 10/3

3a + 3ar^2 = 10r

3r^2 – 10r + 3 = 0

3r^2 – 9r – r + 3 = 0

3r(r – 3) – 1(r – 3) = 0

(r – 3)(3r – 1) = 0

r = 3, r = 1/3

now for a = -1 we get

-3 – 3r^2 = 10r

3r^2 + 10r + 3 = 0

3r^2 + 9r + r + 3 = 0

3r(r + 3) + 1(r + 3) = 0

3r + 1 = 0 , r + 3 = 0

r = -1/3 , r = -3

Now GP will be 27, 3, and 1 / 3, 1 / 27  

Answer 2

Refer to the attachment above for your answer!!

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