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# In a sequence, each term is obtained by adding 4 to the preceding one

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Manager
Joined: 24 May 2016
Posts: 144
In a sequence, each term is obtained by adding 4 to the preceding one  [#permalink]

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12 Aug 2016, 03:25
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45% (medium)

Question Stats:

76% (02:59) correct 24% (02:20) wrong based on 155 sessions

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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

Intern
Joined: 11 Jan 2013
Posts: 5
Re: In a sequence, each term is obtained by adding 4 to the preceding one  [#permalink]

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12 Aug 2016, 06:13
1
1
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
CEO
Joined: 11 Sep 2015
Posts: 3334
Re: In a sequence, each term is obtained by adding 4 to the preceding one  [#permalink]

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12 Aug 2016, 07:29
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Top Contributor
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EBITDA 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

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

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Joined: 26 Nov 2012
Posts: 592
Re: In a sequence, each term is obtained by adding 4 to the preceding one  [#permalink]

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13 Aug 2016, 13:51
1
2
EBITDA 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

a is first term and d = 4 since the preceding number is always greater than the previous one..

Now 10th term is a+9(4).

Sum of n terms is n/2(first term + last term) and here we have to get sum of 10 terms , we get 10/2 ( a + a+36) = 80 ( given sum of 10 terms is 80)

=> 5(2a+36) = 80
=> 2a+36 = 16
=> a = -10

Now to get sum of first 40 terms , we need to get 40th term value i.e. -10+39(4) = 146.

Sum of first 40 terms = 40/2( -10 + 146) = 2720.

Hence option C is correct answer..
Intern
Joined: 17 Nov 2016
Posts: 26
In a sequence, each term is obtained by adding 4 to the preceding one  [#permalink]

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15 Feb 2017, 08:19
Quote:
Sum of AP = n/2 (first term +last term)

Can we apply this formula on any sequence?

Thanks
Zoser
Retired Moderator
Joined: 26 Nov 2012
Posts: 592
Re: In a sequence, each term is obtained by adding 4 to the preceding one  [#permalink]

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15 Mar 2017, 02:46
Zoser wrote:
Quote:
Sum of AP = n/2 (first term +last term)

Can we apply this formula on any sequence?

Thanks
Zoser

Hi Zosar,

Sum of AP = n/2 (first term +last term) - This is only used only for Arithmetic progressions.

For Geometric progression - a *( r^n - 1 / r -1 ) - where a is first term and r is the ration.

For ex: we have series 3,9,27 etc... and we want to find the 10th sum...then a = 3 , r = 3 and n = 10 and use the formula.

Hope this clears..
Intern
Joined: 08 Feb 2017
Posts: 15
Re: In a sequence, each term is obtained by adding 4 to the preceding one  [#permalink]

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20 Mar 2017, 09:45
Jus calculate the 1st series
10x+180(sum of 9 multiples of 4)=80
x=-10
Last number in the series=36+-10=26

It will have total 10 number of 26 i.e. 26*10=260
Rest 10 multiples of 4 i.e. from 4 to 40
Sum of 1st 10 multiples of 4=220
Total = 260+220=480

3rd series will begin with 66(26+40)
66*10=660
660+220=880

4th series will begin with 106(66+40)
106*10=1060
1060+220=1280

Total=80+480+880+1280=2720

option "c"
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Location: India
Re: In a sequence, each term is obtained by adding 4 to the preceding one  [#permalink]

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21 Mar 2017, 02:50
EBITDA 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

sum = 10/2 . (2a +9*4)

or 80 = 5*(2a +36)

or 16 = 2a +36

or 8 = a + 18 or a = -10

sum of 40 terms = 40/2 . (-10*2 + 9*39) = 2720
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Re: In a sequence, each term is obtained by adding 4 to the preceding one  [#permalink]

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17 Sep 2018, 02:19
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