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12 Aug 2016, 04:25
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
77% (02:51) correct
23%
(02:31)
wrong
based on 435
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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
Please explain in detail the answer.
A) 94
B) 320
C) 2,720
D) 27,200
E) 54,400
Please explain in detail the answer.
12 Aug 2016, 07:13
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The question says its an AP with difference of 4 between each term.
Sum of AP = n/2 (first term +last term)
80=10/2(a+l)
hence a+l = 16 (eq 1)
Last term of ap given by
l=a+(n-1)d
l=a + (9)4
a-l =36 (eq 2)
Solving eq 1 and 2 simultaneously you get a=-10 and l=26
Now to find the sum of 1st 40 terms , find the last term
l=a+ (n-1)d
l=-10+39*4
l=146
Now finally find the sum
s= n/2(a+l)
s=40/2(-10+146)
s=20*136
s=2720
Hence answer is C
Sum of AP = n/2 (first term +last term)
80=10/2(a+l)
hence a+l = 16 (eq 1)
Last term of ap given by
l=a+(n-1)d
l=a + (9)4
a-l =36 (eq 2)
Solving eq 1 and 2 simultaneously you get a=-10 and l=26
Now to find the sum of 1st 40 terms , find the last term
l=a+ (n-1)d
l=-10+39*4
l=146
Now finally find the sum
s= n/2(a+l)
s=40/2(-10+146)
s=20*136
s=2720
Hence answer is C
Updated on: 10 Jan 2020, 06:43
Originally posted by BrentGMATPrepNow on 12 Aug 2016, 08:29.
Last edited by BrentGMATPrepNow on 10 Jan 2020, 06:43, edited 1 time in total.
Last edited by BrentGMATPrepNow on 10 Jan 2020, 06:43, edited 1 time in total.
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EBITDA
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:
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