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In a certain sequence, the term an is defined by the formula a_n =

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In a certain sequence, the term an is defined by the formula a_n =  [#permalink]

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New post 29 Jul 2018, 23:40
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In a certain sequence, the term an is defined by the formula \(a_n = 2 *a_{n - 1}\) for each integer n ≥ 2. If \(a_1 = 1\), what is the positive difference between the sum of the first 10 terms of the sequence and the sum of the 11th and 12th terms of the same sequence?

(A) 1
(B) 1,024
(C) 1,025
(D) 2,048
(E) 2,049

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Re: In a certain sequence, the term an is defined by the formula a_n =  [#permalink]

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New post 30 Jul 2018, 01:51
Bunuel wrote:
In a certain sequence, the term an is defined by the formula \(a_n = 2 *a_{n - 1}\) for each integer n ≥ 2. If \(a_1 = 1\), what is the positive difference between the sum of the first 10 terms of the sequence and the sum of the 11th and 12th terms of the same sequence?

(A) 1
(B) 1,024
(C) 1,025
(D) 2,048
(E) 2,049


This tells us that each term is twice of previous term..
So sum of first 10 terms = \(1+2+2^2+2^3+........2^9\)
If you observe each term is ONE more than the sum of previous term that is 2 is 1 more than 1..
2^2 is 1 more than 1+2
2^3 is 1 more than 1+2+2^2 and so on..
Therefore 2^10, the 11th term will be 1 more than sum of first 10 numbers in sequence..
Our answer is therefore \(T_12+T_11-S_10=2^11+1=2049\)
E..

Ofcourse you can do it with geometric progression..
\(S_10=1+2+2^2+....+2^9\)
Now geometric progression formula is [\(fraction]a(r^n-1)/(r-1)[/fraction]=1*(2^10-1)\)
Sum of 11th and 12th term =\(2^10+2^11\)

Difference=\(2^10+2^11-(2^10-1)=2^10+2^11-2^10+1=2^11+1=2049\)
E
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: In a certain sequence, the term an is defined by the formula a_n = &nbs [#permalink] 30 Jul 2018, 01:51
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