JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Application +4 | -1

(1+3+5+\ldots+p)+(1+

3+5+\ldots+q)=(1+3+5+\ldots+r)

where each set of parentheses contains the sum of consecutive odd integers as shown, the smallest possible value of

p+q+r

, (where

p>6

) is

View answer and explanation

Correct option: B

Option B (21) is correct because the problem can be reduced to finding a Pythagorean triplet, and the smallest possible value of p+q+r is 21.

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