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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## There is a sequence A(n) where n is a positive integer... tagged by: swerve ##### This topic has 1 expert reply and 0 member replies ### Top Member ## There is a sequence A(n) where n is a positive integer... There is a sequence A(n) where n is a positive integer such that A(n+1) = 10 + 0.5A(n). Which of the following is closest to A(1,000)? A. 15 B. 18 C. 20 D. 25 E. 50 The OA is C. Please, can any expert explain this PS question for me? I can't get the correct answer. I need your help. Thanks. ### GMAT/MBA Expert Legendary Member Joined 20 Jul 2017 Posted: 503 messages Followed by: 10 members Upvotes: 86 GMAT Score: 770 Let's try building out the first few integers in our sequence to see if we notice a pattern: A(n+1) = 10 + 0.5A(n) A(1) = A(0+1) --> Note: We can't solve for this because we need the value for A(0), which doesn't exist. A(2)= A(1+2) $$=10+\frac{1}{2}A\left(1\right)$$ A(3) = A(2+1) $$=10+\frac{1}{2}A\left(2\right)=10+\frac{1}{2}\left(10+\frac{1}{2}A\left(1\right)\right)=10+5+\frac{1}{4}A\left(1\right)=15+\frac{1}{4}A\left(1\right)$$ A(4) = A(3+1) $$=10+\frac{1}{2}A\left(3\right)=10+\frac{1}{2}\left(15+\frac{1}{4}A\left(1\right)\right)=15+2.5+\frac{1}{8}A\left(1\right)=17.5+\frac{1}{8}A\left(1\right)$$ A(5) = A(4+1)$ $$=10+\frac{1}{2}A\left(4\right)=10+\frac{1}{2}\left(17.5+\frac{1}{8}A\left(1\right)\right)=17.5+1.25+\frac{1}{16}A\left(1\right)=18.75+\frac{1}{16}A\left(1\right)$$

At this point (or maybe a couple numbers back), we can probably recognize a pattern - for each number we increase, we add 10/(2^n-1) and reduce A(1) by 1/2. Knowing this, we can extend our pattern further:

$$A\left(6\right)=18.75\ +\ 0.625\ +\ \frac{1}{32}A\left(1\right)=19.375+\frac{1}{32}A\left(1\right)$$ $$A\left(7\right)=19.375\ +\ 0.3125\ +\ \frac{1}{64}A\left(1\right)=19.6875+\frac{1}{64}A\left(1\right)$$ $$A\left(8\right)=19.6875\ +\ 0.15625\ +\ \frac{1}{128}A\left(1\right)=19.84375+\frac{1}{124}A\left(1\right)$$

Now we should notice that 1) Our fraction of A(1) is becoming very small very fast, and 2) our number is increasing slower and slower as it approaches 20. Knowing this, we can extrapolate that by the time we reach A(1000), A(1) will be such a small fraction that it will be negligible, and our number will still be close to 20. So A(1000) will be closest to 20, or answer choice C.

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