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Apr 2808:00 AM PDT
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30 Jul 2018, 00:40
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E
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Difficulty:
65%
(hard)
Question Stats:
68% (02:54) correct
32%
(03:00)
wrong
based on 474
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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
(A) 1
(B) 1,024
(C) 1,025
(D) 2,048
(E) 2,049
30 Jul 2018, 02:51
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Bunuel
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
13 Feb 2020, 00:55
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Bunuel
Solution
- • \(a_n = 2* a_{(n-1)}\) for integers n≥2
- o or, \(\frac{a_n}{a_{(n-1)}} = 2\)
o Thus, sequence is a geometric progression with \(a_1 =1\) and common ratio (let’s say r) as 2.
- And, nth term in the sequence will be, \(a_n = a_1*2^{(n-1)} = 2^{(n-1)}\)
• Now, the sum of 11th term and 12th term \(= a_{11} +a_{12} = 2^{10} + 2^{11}…………Eq. (2)\)
• Subtracting Eq. (1) from Eq.(2), we will get the required positive difference.
- o Hence, Required positive difference \(= 2^{10}+2^{11} – 2^{10} +1 = 2^{11} +1 = 2048+1 = 2049\)
