[Camellia Ray]

N/10= 1-N/10 at N =5; so when the frog  is aat lily pad 5 it has equal probability of either being eaten up by the snake or get escaped. Let (P_k) represents the probability that the frog escapes if it is currently on pad k then (P_5)= 1/2.

Solving the following equations

(P_1)= 9/10 (P_2)

(P_2)= 2/10 (P_1)+8/10 (P_3)

(P_3)= 3/10(P_2)+ 7/10 (P_4)

(P_4)= 4/10 (P_3)+6/10 (P_5)

(P_5)= 1/2

SOLVING THE ABOVE EQUATIONS STEP BY STEP WE GET

\(P_4\)= 2/5 \(P_3\)+ 3/10

(P_3)= 3/10 (P_2) +7/10 ×(2/5 (P_3 +3/10))

(P_3)= 3/10 (P_2) + 14/50 (P_3) +21/100

36/50 (P_3)= 3/10 (P_2) + 21/100

\P_3)= (3/10 (P_2) +21/100)×50/36

(P_2)= 2/10 (P_1) + 8/10 ( 3/10 (P_2) +21/100)×50/36

= 2/10 (P_1) +1/3 (P_2) + 7/30

2/3 (P_2)=2/10 (P_1)+7/30

(P_2) = 3/10 (P_1) + 7/20

(P_1) = 9/10×(3/10 (P_1) + 7/20)

(P_1)= 27/100 (P_1) +63/200

(P_1)×73/100= 63/200

(P_1)= 63/146

 

 

 

 

 

 

 

 

[Camellia Ray]

In the given question it has been asked to find the no of subsets which satisfy the condition that
A={a, a+d,a+2d….a+kd} &

A U{x}  is no longer an A.P where x ∈ (S-A)

From above 2 condns it can b concluded that at a time with c.d d we have to form the largest sequence having c.d d and starting element say a

For ex if X={1,2,3,,,,12} then for

c.d 1:  only 1 subset possible; c.d 2 ->2 :subset possible namely { 1,3,5,7,,,11} & {2,4,6,8,,,,12}

Now c.d 1 and c.d 11 only 1 subset; c.d 2 &10 ->2 subsets

GENERALIZATION

For N Odd

1. c.d 1 & n-1 -> 1 subset;  2. c.d 2 & n-2 -> 2 subsets

similarly for c.d (n-1)/2) & cd (n+1)/(2) it will be (n-1)/2 subsets

SO TOTAL IS 2× (1+2+3+…..+(n-1)/2)

FOR EVEN

SAME as previous just one more term n/2 added

so here it is 2×(1+2+….(n−1)/2)+n/2

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