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What is the sum of the series 2, 4, 6, 8, 10...48.  I never seem to get the right...

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susan1988 | Student, Undergraduate | (Level 1) Honors

Posted October 1, 2010 at 12:03 AM via web

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What is the sum of the series 2, 4, 6, 8, 10...48.  I never seem to get the right answer.

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william1941 | College Teacher | (Level 3) Valedictorian

Posted October 1, 2010 at 12:05 AM (Answer #1)

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Just follow these steps carefully and understand what is being done. That way you will not have any problems in the future.

First, for the given series we see that the difference between any term and the one following it is a constant. Therefore the given series in an AP. Now for an AP the nth term is given by a+(n–1)d. Also, the sum of the first n terms is [2a + (n-1)d]* (n/2).

For the given series 48 = 2 + (n-1)*2

=> n-1 = (48 – 2) /2  =  46/2 = 23

=> n = 24

So we need to find to the sum of 24 terms where 2 is the first term and the common difference is 2.

So the sum is [2a + (n-1)d]* (n/2)

=> [2*2 + (24-1)2]* (24/2)

=> [4 + (23)2]* 12

=> (4+ 46)*12

=> 50 * 12

=> 600

Therefore the required sum is 600.

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susan1988 | Student, Undergraduate | (Level 1) Honors

Posted October 1, 2010 at 12:33 AM (Answer #2)

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I can't understand who is right, why has 1200 been given as the answer when 50*24/2= 600

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted October 1, 2010 at 2:19 AM (Answer #3)

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We can see that if we'll calculate the difference between 2 consecutive terms of the given series, we'll obtain the same value each time:

4 - 2 = 6 - 4 = 8 - 6 = 10 - 8 = ...... = 2

So, the given series is an arithmetic progression whose common difference is d = 2.

Now, we can calculate the sum of n terms of an arithmetic progression in this way;

Sn = (a1 + an)*n/2

a1 - the first term of the progression

a1 = 2

an - the n-th term of the progression

an = 48

n - the number of terms

 We can notice that we know the first and the last terms but we don't know the number of terms. We can calculate the number of terms using the formula of general term.

an = a1 + (n-1)*d

48 = 2 + (n-1)*2

We'll remove the brackets:

48 = 2 + 2n - 2

We'll eliminate like terms:

48 = 2n

We'll divide by 2:

n = 24

So, the number of terms, from 2 to 40 is n = 24 terms.

S24 = (2 + 48)*24/2

S24 = 50*12

S24 = 600

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phucnguyen08 | Student, College Freshman | (Level 1) Honors

Posted October 1, 2010 at 10:25 PM (Answer #4)

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from 0 to 10 we have 5 pairs (0+10)+(2+8)+(4+6)=30

Like that form 10-20 we have (10+20)+(12+18)+(14+16)=90

form 30 to 40 we have (30+40)+(32+18)+(34+36)=210

form 40to 46 we have (48+42)+(46+44)=100

The total is 30+90+210+100=430

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neela | High School Teacher | (Level 3) Valedictorian

Posted October 1, 2010 at 12:18 AM (Answer #5)

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The sries 2 , 4 , 8 , ... 48 is a n arithmetic progression, with starting term a1 = 2 and  last term an = 48 and the common diffrence d  between the succesive terms is d =  4-2 = 8-4 = 10-8 = 2.

S the number of terms n is given by:  n = (an -a1 )/d + 48-2)2 +1 = 23+1 = 24.

Therefore the sum Sn = (a1+an)*n/2 = (2+48)*24/2  = 1200

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