IIT JEE , AIEEE Forum - Mathematics , Algebra

Arjun

here i go...........

so since on da first day.....i medal i medal and 1/7 of da remeanin medals......means 6/7medals which wer remainin......after....the day b4 da first day......

this happens everyday...........

on da first da 1+x/7.......medals were given........

on da second day.....2+1/7(x-1-x/7)......here x is da total number of medals.........

so we get.......2+1/7(6x-7/7)

generalising......we get on the pth day.......

m₁=x m_p+1= 6(m_p-p)/p.....

using dis formula and sbstituting values.....

we get an A.G. series ...

x= 1+2(7/6)+3(7/6)².............n(7/6)^n-1

using da formula for summation and upon simplifyin...

we get the sum as.....

m=36(1-(n+1)(7/6)ⁿ+n(7/6)ⁿ⁺¹)

m=36+(n-6)7ⁿ/6^n-1

6 does not divide 7.....for dis part to b a whole number.....

so we get n=6......to make m an integer..........

since n=6.....m=36.........

is da answer....correct.....it involved a lotta tireless......work.......

thank u....

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