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Recursive Sequence Problem!

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Recursive Sequence Problem!

by Duaabasheer » Fri Nov 16, 2012 11:28 am
Hello there!

I was going through the MGMAT Flashcards and for Flashcard 4 of Quant EIVs I really could not understand the solution

Q: What is the 25th term of this sequence? Sn=Sn-1 -10 and S3=0

A: First we need to convert the recursive sequence definition provided into a direct sequence formula. Each term is 10 less than the previous one ; therefore Sn= -10n +k ... and then they solve for S2 and S1 using this sequence formula.

My question is : Why is the formula -10N?? we are not multiplying by 10, rather subtracting 10 from the previous term. I would have thought the formula would be Sn= Sn-1 -10.

Help!

Duaa
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by GMATGuruNY » Fri Nov 16, 2012 12:55 pm
Duaabasheer wrote: What is the 25th term of this sequence? S(n)=S(n-1) -10 and S3=0.
Each term is 10 less than the preceding term.

Since 25-3 = 22, to go from S₃ to S₂₅, 10 must be subtracted 22 times.
Thus:
S₂₅ = S₃ - 22(10) = 0 - 22(10) = -220.
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by Duaabasheer » Fri Nov 16, 2012 10:34 pm
Would have never thought of it that way, thanks!
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by GaneshMalkar » Sat Nov 17, 2012 3:09 am
Sn = Sn-1 - 10

S3 = 0 = S2 - 10 so S2 = 10

S2 = 10 = S1 - 10 so S1 = 20

Sequence is 20,10,0,-10,-20,....

From AP formula = tn = a + (n-1) d
where tn = nth term
a = first term
n = no of terms
d = common diff

t25 = 20 + (25 - 1) (-10) = 20 - 240 = -220 :)
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