if(1+3+5+.....+p)+(1+3+5+...+q)=(1+3+5+...++r) find the minimum value of P + Q + R when p is greater than 6
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7+5+9=21
minimum value of p+q+r=21 after calculate we get (p+1)^2+(q+1)^2=(r+1)^2
as p>6 (said)
so let p=7 and q=5 and also r=9
therefore we have
8^2+6^2=10^2 which is true
minimum value of p+q+r=21 after calculate we get (p+1)^2+(q+1)^2=(r+1)^2
as p>6 (said)
so let p=7 and q=5 and also r=9
therefore we have
8^2+6^2=10^2 which is true
mohitkantewal:
r is p+q
than how it is 9
why r is p+q?
taking the sum of series we get the relation (p+1)^2+(q+1)^2=(r+1)^2 from this relation we have to find the minimun value of p+q+r. also given that p>6....i hope you got it
p>6
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