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In a sequence, each term is obtained by adding 4 to the preceding one.
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11 May 2021, 03:48
11 May 2021, 03:48
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In a sequence, each term is obtained by adding 4 to the preceding one. If the sum of the first 10 terms is equal to 80, what is the result of the addition of the first 40 terms?
A) 94
B) 320
C) 2,720
D) 27,200
E) 54,400
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
A) 94
B) 320
C) 2,720
D) 27,200
E) 54,400
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: In a sequence, each term is obtained by adding 4 to the preceding one.
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11 May 2021, 05:53
11 May 2021, 05:53
Carcass wrote:
In a sequence, each term is obtained by adding 4 to the preceding one. If the sum of the first 10 terms is equal to 80, what is the result of the addition of the first 40 terms?
A) 94
B) 320
C) 2,720
D) 27,200
E) 54,400
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
A) 94
B) 320
C) 2,720
D) 27,200
E) 54,400
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Here's another solution that requires only 1 formula: Sum of the integers from 1 to n inclusive = (n)(n+1)/2
Let a = term1
So, the sequence looks like this:
term1 = a
term2 = a + 4
term3 = a + 4 + 4
.
.
.
term10 = a + + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4
The sum of the first 10 terms is equal to 80
So: 80 = (a) + (a + 4) + (a + 4 + 4) + .... + (a + + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4)
80 = 10a + a bunch of 4's
How many 4's?
(1 four) + (2 fours) + (3 fours) + . . . + (9 fours)
Applying the formula, the sum of the integers from 1 to 9 = (9)(10)/2 = 45. So, there are 45 4's in the sum.
We get: 80 = 10a + (45)(4)
Simplify: 80 = 10a + 180
We get: -100 = 10a
Solve: a = -10. Great, the first term is -10
What is the sum of the first 40 terms?
Our sequence is as follows:
term1 = -10
term2 = -10 + 4
term3 = -10 + 4 + 4
term4 = -10 + 4 + 4 + 4
.
.
.
term40 = -10 + (sum of 39 4's)
So, the sum of the first 40 terms = -400 + a bunch of 4's
How many 4's are there altogether?
(1 four) + (2 fours) + (3 fours) + . . . + (39 fours)
Applying the formula, the sum of the integers from 1 to 39 = (39)(40)/2 = 780
So, the sum of the first 40 terms = -400 + (780)(4) = 2720
Answer: C
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Re: In a sequence, each term is obtained by adding 4 to the preceding one.
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08 Jul 2021, 06:43
08 Jul 2021, 06:43
Theory: Arithmetic Sequence: A sequence of numbers such that the difference between the consecutive term is constant
Sum of n terms \(S_n\) of an Arithmetic sequence is given by \(S_n\) = \(\frac{n}{2}\) * (2a + (n-1)*d)
where, a is the first term
n is number of terms or term number
d is the common difference (Difference between consecutive terms)
Given series has common difference, d as 4 [ Because each term is obtained by adding 4 to the preceding one]
Sum of first 10 terms, \(S_{10}\) = 80 [Given]
Using \(S_n\) formula we get
\(S_{10}\) = \(\frac{10}{2}\) * (2a + (10-1)*d) = \(\frac{10}{2}\) * (2a + 9*4)
= 5*(2a + 36) = 80
=> 2a + 36 = \(\frac{80}{5}\) = 16
=> 2a = 16 - 36 = -20
=> a = \(\frac{-20}{2}\) = -10
We need to find \(S_{40}\)
\(S_{40}\) = \(\frac{40}{2}\) * (2a + (40-1)*d) = \(\frac{40}{2}\) * (2*-10 + 39*4)
= 20*(-20+156) = 20*136= 2720
So, answer will be C
Hope it helps!
To Learn More about Arithmetic or Geometric Sequence watch the following video
Sum of n terms \(S_n\) of an Arithmetic sequence is given by \(S_n\) = \(\frac{n}{2}\) * (2a + (n-1)*d)
where, a is the first term
n is number of terms or term number
d is the common difference (Difference between consecutive terms)
Given series has common difference, d as 4 [ Because each term is obtained by adding 4 to the preceding one]
Sum of first 10 terms, \(S_{10}\) = 80 [Given]
Using \(S_n\) formula we get
\(S_{10}\) = \(\frac{10}{2}\) * (2a + (10-1)*d) = \(\frac{10}{2}\) * (2a + 9*4)
= 5*(2a + 36) = 80
=> 2a + 36 = \(\frac{80}{5}\) = 16
=> 2a = 16 - 36 = -20
=> a = \(\frac{-20}{2}\) = -10
We need to find \(S_{40}\)
\(S_{40}\) = \(\frac{40}{2}\) * (2a + (40-1)*d) = \(\frac{40}{2}\) * (2*-10 + 39*4)
= 20*(-20+156) = 20*136= 2720
So, answer will be C
Hope it helps!
To Learn More about Arithmetic or Geometric Sequence watch the following video
