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

Question

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

chetan2u · Answer

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) =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

GMATWhizTeam · Answer

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)}    
• Sum of first 10 terms  = \frac{a_1(r^ {10}-1)}{(r-1)} = 2^ {10} -1 ……….Eq.(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    
Thus, the correct answer is  Option E.

CodeBXT · Answer

Solution: This is a simple way to get it done.

Given: an=2*a(n−1) for integers n≥2
a1=1

So, using an=2*a(n-1) we have,

a2= 2*a1 = 2*1=2
a3=2*2=4
a4=2*4=8

We can see the sequence progressing in geometric progression (GP) with a common ration (r)= 2. So using the formula for GP, we can find the 10th term,
i.e. a10=a1*r power (n-1) 
 = 1* 2power9 =1*512= 512

similarly, a11= 1*2power10= 1024
 a12= 1*2power11= 2048

Then the sum of first ten terms can be found using Sum for Geometric Progression, remember since this is geometric progression, we cannot use sum= avg*no. of terms. Rather we have to use,

Sum (Sn)= a1*(rpowern - 1)/ r - 1

So S10= 1* (2power10 - 1)/2-1
 = 1024-1/1
 =1023

Hence, the 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 is 
a11+a12- Sum (S10) = 1024+2048-1023 =2059.

Hence, answer choice E is the correct answer.

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

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