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# If (1+3+5+.....+p) + (1+3+5+.....+q) = (1+3+5+.....+r) where each set of parentheses contains the sum of consecutive integers as shown, the smallest possible value of p+q+r, (where p>6) is(a) 12(b) 21(c) 45(d) 54

SAGAR SINGH - IIT DELHI
879 Points
9 years ago

Dear student,

The correct option is c)

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Sagar Singh - IIT Delhi

iit jee
14 Points
8 years ago

ans is 21.

i''ll show u how...--->

no. of terms--> (p+1)/2, (q+1)/2, (r+1)/2

sum of 1st n odd nos. =n^2

((p+1)/2)^2 +  ((q+1)/2)^2 =  ((r+1)/2)^2

(p+1)^2 + (q+1)^2 = (r+1)^2

p+1,q+1,r+1 are pythagorean triplets

p>6...........p+1>7

therfore the smallest set is (8,6,10)......p+1=8,q+1=6,r+1=10................p+q+r=21......ans....simple