Question

If (1 + 3 + 5 + ... + p) + (1 + 3 + 5+ ... +q) = (1 + 3 + 5 + ... +r) where each set of parenthesis contains
the sum of consecutive integers, then the smallest possible value of p+q+r(p >6) is​

Answer 1

Answer:

1+3+5...........+2n - 1

no. of terms = n + 1 / 2

according to question we can write :

[p + 1/2 ]^ 2 + [ q + 1/2 ]^2 = [ r + 1/2 ]^2

IT form Pythagorean triplet thus for min value we takes 3, 4, 5, as of our Pythagorean triplet , therefore ,

p + 1/2 = 4 , q + 1/2 = 3 , r +1/2 = 5

p=7 , q = 5 , and r = 9

p + q + r = 21

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Answer 2

Answer:

see this attachment....

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