Question

From three numbers in Gp other three numbers in Gp are subtracted and the remainder are found in GP. Prove that the three sequence 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.....