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maxpowers
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What is the sum of 1+3+5...+71+73+75 14 Feb 2007, 15:34
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What is the sum of 1+3+5...+71+73+75?



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What is the sum of 1+3+5...+71+73+75 14 Feb 2007, 17:51
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What is the sum of 1+3+5...+71+73+75?

arithmatic sequnece

d = 2 , n= 75 , average = 38
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What is the sum of 1+3+5...+71+73+75 15 Feb 2007, 13:39
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yezz
What is the sum of 1+3+5...+71+73+75?

arithmatic sequnece

d = 2 , n= 75 , average = 38
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What is your answer? I don't understand what you're donig with d, n, and the average. I solved it with a different method.
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What is the sum of 1+3+5...+71+73+75 15 Feb 2007, 15:00
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maxpowers
What is the sum of 1+3+5...+71+73+75?
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Sum of Arithmatic progression

a + (a+d) + (a+2d) .... + (a+(N-1)d))
where a is 1st term d is diffrence b/w 2nd and 1st
and total N terms

Sum = (N/2)[2*a+(N-1]*d)

How many terms are there using Nth term = a + (N-1)*d

last term 75=1+(N-1)*2==> N=38

There are 38 terms

Sume = (38/2)( 2*1+(37)*2)= 1444
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What is the sum of 1+3+5...+71+73+75 15 Feb 2007, 15:39
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1+75 = 76
3+73 = 76
5+71 = 76 etc

If you would have a sequence of distinct integers from 1 to 75, there would be (75+1)/2 = 38 pairs

Because there is every second integer in a sequence, you have 38/2 = 19 pairs

19*76 = 1444
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What is the sum of 1+3+5...+71+73+75 16 Feb 2007, 05:13
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Or simply use Average formula.

(75+1)/2=38

Number of terms 75-1=74/2=37+1=38

38*38=1444
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What is the sum of 1+3+5...+71+73+75 16 Feb 2007, 08:57
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The answer is 1444. How do you figure out the number of terms?
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What is the sum of 1+3+5...+71+73+75 16 Feb 2007, 10:52
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How many terms are there using formula
Nth term = a + (N-1)*d
where a is 1st term and d is differnce and N is number of terms

last term 75=1+(N-1)*2==> N=38

There are 38 terms
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What is the sum of 1+3+5...+71+73+75 17 Feb 2007, 04:50
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sorry if i was not clear , i was trying to give u hints , so u can do it yourself

u can solve it mathematical progression or usinge the average \9 middle term* to number of terms



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