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(1+3+5+.....+p)+(1+3+5........+q)=(1+3+5+.....+r) where p q r are real p greater than 6 what is the smallest possible value of p+q+r?

(1+3+5+.....+p)+(1+3+5........+q)=(1+3+5+.....+r)


where p q r are real p greater than 6


what is the smallest possible value of p+q+r?

Grade:12

2 Answers

bhanuveer danduboyina
95 Points
10 years ago

1+3+5+......p=p/2[p+(p-1)2]

1+3+5+......q=q/2[q+(q-1)2]

1+3+5+......r=r/2[r+(r-1)2]

p/2[p+(p-1)2] + q/2[q+(q-1)2] = r/2[r+(r-1)2]

On simplification,

p2 + q2 = r2

Given:

p>6 ; On further simplification,

we get p=8 , q=6 , r=10 are the smallest values

=> p + q + r = 24

ujjwal dubey
14 Points
10 years ago

actually the answer is cumin 21

(book: tata mc graw hill iit jee - 2010)

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