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Series - Arithmetic Progression (1 Viewer)

cookiez69

What a stupid name, Nat.
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Hey guys,

Two questions,

Q: Find the sum of: 1 - 2 + 3 - 4 + 5 -6 + ... - 100

Q: What is the least number of terms required if the sum of 15 + 20 + 25 +... is to exceed 2625?

Thanks in advance!
 

Makematics

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for the first one split it into (1+3+5+... +99) - (2+4+6+...+100) and use the sum of an AP
 

Sy123

This too shall pass
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Hey guys,

Two questions,

Q: Find the sum of: 1 - 2 + 3 - 4 + 5 -6 + ... - 100

Q: What is the least number of terms required if the sum of 15 + 20 + 25 +... is to exceed 2625?

Thanks in advance!
First one:

We cannot deal with that series as a whole, so its best to split it into 2 separate ones:



Each of them are arithmetic series each with a common difference of 2 (but different initial term). So using the sum formula:



Second one:

We will denote a general term:

We need to find the lowest k required so that:



Sum the LHS using the sum formula:









Now we simply have to solve it.



(using the quadratic formula I found the 'roots')

And since k must be positive:



That means the lowest value of k, or the least number of terms required is 31. (not 30 as that equals 2625).
 

RealiseNothing

what is that?It is Cowpea
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For the first question, instead of splitting it into 2 sums, you can just simplify:





etc



So you are just adding up 50 lots of -1 to get -50.
 

RealiseNothing

what is that?It is Cowpea
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Q: What is the least number of terms required if the sum of 15 + 20 + 25 +... is to exceed 2625?
Let's add 5 + 10 to each side:

"What is the least number of terms required if the sum of 5 + 10 + 15 + 20 + 25 +... is to exceed 2640?"

We can divide everything by 5 to obtain:

"What is the least number of terms required if the sum of 1 + 2 + 3 + 4 + 5 +... is to exceed 528?"

Now we just find the first triangular number that exceeds 528:









But we added two extra terms by adding the '5 + 10', so:



Thus we need 31 terms.
 

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