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finding the sum of consecutive integers

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finding the sum of consecutive integers

by fangtray » Wed May 09, 2012 9:05 pm
I would like to know if the following ways are correct and if there is an easier way to solve:

how do you find the sum of the first consecutive positive even integers?

2n+2 = 2(50)+2 = 102 so the 50th consecutive integer. the first would be 2
the sum would be the average * the number of terms.
so 102+2/2 = 52 * 50

how do you find the sum of the first consecutive postiive odd integers?
similiarly,
2n+1 for odd integers..so 101 is the 50th consecutive odd integer.

101+1 = 102/2 so 51 * 50 would be the sum..


What if we want to find the sum of all consecutive positive integers from 17-401?

What if we want to find the sum of all consecutive multiples of 3 from 17-401?

is there a way to do this using the strategy above? is there a simpler way?
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by aneesh.kg » Wed May 09, 2012 9:27 pm
What if we want to find the sum of all consecutive positive integers from 17-401?
To find the sum of this AP, we need the number of terms. How do we find that?
AP comes to the rescue again!

The number of terms can be found by this formula of general term of an AP:
an = a + (n - 1)*d

If an is the last term, then n is the number of terms in this AP.
Constant Difference = 1 for consecutive integers.
so,
401 = 17 + (n - 1)*1
384 = n - 1
n = 385

Sum
= n/2 * (a + l)
= 385/2*(17 + 401)
= 385*209
What if we want to find the sum of all consecutive multiples of 3 from 17-401?
Same strategy, only that the constant difference = 3 this time.
an = a + (n - 1)d
401 = 17 + (n - 1)*3
n - 1 = 384/3
n = 129

Sum
= 129/2 * (17 + 401)
= 129*209
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by fangtray » Fri May 11, 2012 3:27 pm
aneesh.kg wrote:
What if we want to find the sum of all consecutive positive integers from 17-401?
To find the sum of this AP, we need the number of terms. How do we find that?
AP comes to the rescue again!

The number of terms can be found by this formula of general term of an AP:
an = a + (n - 1)*d

If an is the last term, then n is the number of terms in this AP.
Constant Difference = 1 for consecutive integers.
so,
401 = 17 + (n - 1)*1
384 = n - 1
n = 385

Sum
= n/2 * (a + l)
= 385/2*(17 + 401)
= 385*209
What if we want to find the sum of all consecutive multiples of 3 from 17-401?
Same strategy, only that the constant difference = 3 this time.
an = a + (n - 1)d
401 = 17 + (n - 1)*3
n - 1 = 384/3
n = 129

Sum
= 129/2 * (17 + 401)
= 129*209
whoa i don't know this formula. it looks useful. what does ap stand for? and will it work with ALL consecutive integer problems?
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by aneesh.kg » Fri May 11, 2012 9:49 pm
Ok, let me tell you briefly about
Arithmetic Progression (AP)

Lets take a few sequences of numbers
6,7,8,9,10,11
-3,-1,1,3,5
8,5,2,-1,-4

Do you notice a pattern in the sequences above?
The difference between any two consecutive terms in each of those sequences is Constant throughout the sequence. This difference is called Constant Difference (d) and such a sequence is called an Arithmetic Progression (AP).

If the first term of the AP is a, the second term is (a + d), the third term is (a + 2d)..
and the nth term is?
a + (n - 1)d = nth term

Also, Sum of an AP = (n/2)*(first term + last term) = (n/2)*(2a + (n - 1)*d)

All consecutive integers are an AP with constant difference (d) = 1
All consecutive odd integers or even integers are an AP with constant difference (d) = 2
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by fangtray » Mon May 14, 2012 2:56 pm
aneesh.kg wrote:Ok, let me tell you briefly about
Arithmetic Progression (AP)

Lets take a few sequences of numbers
6,7,8,9,10,11
-3,-1,1,3,5
8,5,2,-1,-4

Do you notice a pattern in the sequences above?
The difference between any two consecutive terms in each of those sequences is Constant throughout the sequence. This difference is called Constant Difference (d) and such a sequence is called an Arithmetic Progression (AP).

If the first term of the AP is a, the second term is (a + d), the third term is (a + 2d)..
and the nth term is?
a + (n - 1)d = nth term

Also, Sum of an AP = (n/2)*(first term + last term) = (n/2)*(2a + (n - 1)*d)

All consecutive integers are an AP with constant difference (d) = 1
All consecutive odd integers or even integers are an AP with constant difference (d) = 2

wow thanks aneesh! when solving for n, will you ever get a decimal?

for example the formula 401 = 17 + (n-1)3 N = 129. but will N ever be a decimal? or if you had to solve for AN or A?
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by aneesh.kg » Mon May 14, 2012 9:48 pm
Hi,

It's good to see you thinking about this concept.

'n' will always be a positive integer because it represents the number of terms in the AP.

if instead of
401 = 17 + (n-1)3, you have
400 = 17 + (n-1)3, and get
n = 128.66, then you should realise that in such an AP, it is impossible to have 401 as the last term.

'n' will not be an integer (for an AP) only when:
(i) You've made a mistake somewhere.
(ii) There is a mistake in the problem.

Let me know if there's anything else.
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