Question

what number must be subtracted from each of the numbers 31 26 and 22 so that remainder may be in continuous proportion​

Answer 1

Let x be the number subtracted from each numbers.

Then the numbers are (31-x), (26-x) amd (22-x).

(31-x)=(26-x)=(22-x)

=>31-26-22=x+x-x

=>2x=-17

=>x=-8.5

Hence, -8.5 should be subtracted from each number so that the remainder is proportional.

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Answer 2

Step-by-step explanation:

• Let the number to be subtracted be m.
• So 31 – 3m, 26 – m, 22 – m are in continued proportion.
• Now if m : n : : n : o, then m,n,o are in continued proportion and o is the third proportion of m and n.
• Now we have b^2 = ac
•                   So n^2 = mo

                       (26 – m)^2 = (31 – m) (22 – m)

•             So by using  (a -  b)^2 = a^2 + b^2 – 2ab we get
•                  (26)^2 + m^2 – 2(26)m = 682 – 22 m – 31 m + m^2
•                   676 + m^2 – 52 m = 682 – 53 m + m^2
•                                 m – 6 = 0
•                                 Or m = 6
• Therefore 6 has to be subtracted from 31,26 and 22 so that they are in continued proportion.

Reference link will be

https://brainly.in/question/3111201