IIT JEE past questionAP GP

If thotal number of runds scored in n matches is \left(\frac{n+1}{4} \right)\left(2^{n+1} -n-2\right) where n >1, and the runs scored in the kth match are given by k.2n+1-k, 1≤k≤n.

Find n

3 Answers

1
Optimus Prime ·

total runs made =1.2n+2.2n+.....+n.2=4(2n-1)-2n

4(2n-1)-2n=(n+1)(2n+1-n-2)/4

16(2n-1)-8n=(n+1)2n+1-(n+2)(n+1)

(n+2)(n-7)=(n-7)(2n+1

n=7

2n+1=n+2 has no valid solutions
hence n=7

1
Optimus Prime ·

what isthe correct answer nishant sir?

21
tapanmast Vora ·

Tot. runs = 2n+1Σ(k/2k )

tot runs = 2^n+1 ....... the fol. is an A.G.P series.....

RHS = 2n+1(2 -21-n) + n.2n)

Now compare wid tot. runs scored as given on LHS

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