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

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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?

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

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

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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.

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

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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

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

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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

gmatclubot

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