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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
Dear student,
The correct option is c)
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We are all IITians and here to help you in your IIT JEE preparation.
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Askiitians Expert
Sagar Singh - IIT Delhi
sagar singh....u r mad....
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
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