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Math Expert
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Re: Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined [#permalink]
Definitely agree with the analysis above, but you could also plug in the numbers into the equations for P(n) and Q(n) - I think that will be faster
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Re: Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined [#permalink]
p1 = 11 ; q3 = 103 . So they both are already in form p(k) and q(k+2) .
q3 - p1 = 103-11 = 92 .

Decrease in difference after every iteration to higher term = 7 ( +3 for p series and -4 for q series).
Question is asking for what value of 'K' or after how many such iteration from P1 will q series be lesser than p series ?

Ans = 92-: 7 = 14 i.e after 14 such iteration q series will finally be lesser than p series . After 13 iterations there will still be a difference of 1 (7*13 =91). Remember p series has to be greater than q series. Hence 14.
Value of P1 after 14 iterations = P15 , as 1st iteration gets us the value of P2 not P1.(P1 is already given ).

Answer would have been 'D' if p0 and q2 was given . I hope this will explain the solution.

Ans E
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Re: Sequence P is defined by pn = p(n-1)+ 3, p1= 11, sequence Q is defined [#permalink]
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