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## Formula for AP/GP?

80.1/2+1/4+1/8+...+1/512=?
A.0<S<1/2
B.1/2<S<1
C.1<S<3/2
D. 3/2<S<2
E. 2<S<5/2

Don't know if this is a GP/AP problem. What is the formula for GP/AP. Also what is the difference between GP and AP?

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1/2+1/4+1/8+...+1/512=?....the terms int his sequence are in Geometric progression,
first term a1 = 1/2
r = a(n+1)/a(n) = 1/2
number of terms (n) = 512/2 = 256

Sum or n terms in GP is given by a1(1-r^n)/1-r
= 1/2*[1-1/2^256]/1-1/2 = 1-1/2^256.

Hence, IMO it is B.

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Thanks

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Hi, 1/2 + 1/4 + 1/8 +.... is a GP. If the series extends upto infinity, then the sum is 1. How? For infinite GP,
Sum = a/(1-r); a=1st term & r= ratio of successive terms
=1/2/(1-1/2)=1.

Now the sequence terminates at 1/512. So, the sum must be less than 1. However, the sequence starts at 1/2 and all are positive terms. So the sum must be greater than 1/2. The answer must be B.

NOTE: Although it's not required in this case, you might want to know the actual sum here. Since it's a finite GP, the formula is

SUM =a*(1-r^n)/(1-r)
=1/2*{1-(1/2)^9}/(1-1/2) [1/2+(1/2)^2+(1/2)^3+.....+(1/2)^9 => 9 terms, not 256 terms]
=(1-1/2^9)

Whether you use a*(1-r^n)/(1-r) or a*(r^n-1)/(r-1) does not matter. For the purity of of mathematics, we use the first one when r<1 and the second one when r>1. But all the same result. +/+ or -/-.

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To calculate how many number of terms:
Since, Last Term = First Term * r ^(n-1)
therefore:
1/512 = 1/2 * (1/2)^(n-1)
therefore n = 9

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Now the sequence terminates at 1/512. So, the sum must be less than 1. However, the sequence starts at 1/2 and all are positive terms. So the sum must be greater than 1/2. The answer must be B.

Very nice approach Makumajon

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In a finite geometric progression, S < 1

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Originally Posted by ranjeet_1975
In a finite geometric progression, S < 1
Ranjeet you forgot to mention..only if both the common ratio and first number is a fraction...

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Yes, for example not in the case : 1 + 3+ 9+ 27+......