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4 numbers are in proportion such tht sum of extremes=21 and sum of means = 19. Sum of squares of all 4 numbers =442. How many distinct values can sum of first and third number assume?

4 numbers are in proportion such tht sum of extremes=21 and sum of means = 19. Sum of squares of all 4 numbers =442. How many distinct values can sum of first and third number assume?

Grade:10

1 Answers

Arun
25750 Points
3 years ago
Let the numbers be a,b,c,d
From given a+d=21 and b+c=19
a2+b2+c2+d2=442
As they are in proportion ad=bc
When ratio of 2 terms=ratio of 2 other terms they are in proportion
So (a2+d2+2ad)+(b2+c2+2bc)=441+361
442+4ad=802
ad=90 So ad=bc=90
(a-d)2=(a+d)2-4ad=441-360=81
So a-d=9 and a+d=21
So a=15 d=6
Similarly (b-c)2=361-360=1
So b-c=1 and b+c=19 So b=10,c=9
So Four proportionals are 15,10,9,6

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