Question

From three numbers in GP other three numbers in GP are subtracted and the remainder are also found to be in GP. Prove that the three sequences have the same common ratio.​

Answer 1

Consider the geometric progressions given below :-

a, ar, ar² and b, bn, bn²

Subtracting the individual terms of these GPs, we get :-

(a-b), (ar - bn), (ar² - bn²)

It is given that, this is also a GP

So,

ar - bn = √[(a - b)(ar² - bn²)]

=> (ar - bn)² = (a - b)(ar² - bn²)

=> a²r² + b²n² - 2abrn = a²r² + b²n² - abn² - abr²

=> 2abrn = abn² + abr²

=> 2rn = n² + r²

=> n² - 2rn + r² = 0

=> (n - r)² = 0

=> n = r

So, the GPs will be

a, ar, ar² and b, br, br²

And, their difference :-

(a - b), (a - b)r, (a - b)r²

Thus, all these have a common ratio 'r'.

Hope, it'll help you.....

Answer 2

i hope.it will help you thnx for asking