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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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If $(1+3+5+....+p)$+$(1+3+5+....+q)$ = the smallest value of $p+q+r, (p>6)$ is

$(a)\;54\qquad(b)\;21\qquad(c)\;12\qquad(d)\;45$

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Answer : (b) 21
Explanation : $1+3+5\;.......\;+k=(\frac{k+1}{2})^{2}$
So , $\;(\frac{p+1}{2})^2\;+(\frac{q+1}{2})^2=(\frac{r+1}{2})^2$
$(p+1)^2\;+(q+1)^2\;=\;(r+1)^2$
They form a pythagorean triplet. $\therefore p>6$
$\therefore\;p+1\;>7$
First pythagorean triplet containing a number greater than 7 is $\;(6,8,10).$
$p+1=8\quad\;q+1=6\quad\;r+1=10$
$p+q+r=21.$
answered Jan 20, 2014 by yamini.v
edited Aug 31, 2014 by sharmaaparna1
 

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