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

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

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New post 12 Aug 2016, 04:25
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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.
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Re: In a sequence, each term is obtained by adding 4 to the preceding one [#permalink]

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New post 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
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Re: In a sequence, each term is obtained by adding 4 to the preceding one [#permalink]

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New post 12 Aug 2016, 08:29
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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

Please explain in detail the answer.


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

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New post 13 Aug 2016, 14:51
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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

Please explain in detail the answer.


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

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New post 15 Feb 2017, 09:19
Quote:
Sum of AP = n/2 (first term +last term)


Can we apply this formula on any sequence?

Thanks
Zoser
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Re: In a sequence, each term is obtained by adding 4 to the preceding one [#permalink]

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New post 15 Mar 2017, 03:46
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..
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Re: In a sequence, each term is obtained by adding 4 to the preceding one [#permalink]

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

2nd series will start with 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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Re: In a sequence, each term is obtained by adding 4 to the preceding one [#permalink]

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New post 21 Mar 2017, 03:50
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

Please explain in detail the answer.


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
Re: In a sequence, each term is obtained by adding 4 to the preceding one   [#permalink] 21 Mar 2017, 03:50
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