number sequences....

B

bumblebee123

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I've started on this question, but I can't get passed part a) can anyone help?

question: a) find the sum of the arithmetic series whose first term is 12, common difference is 3 and whose last term is 99
b) find the sum of all 2-digit numbers
c) hence find the sum of all 2-digit numbers which are not multiples of 3

I figured out part a). This was by finding the term number of the last term ( so I would know how many terms are in the series )

99 = a + ( n - 1 )d
99 = 12 + 3n - 3
90 = 3n
n = 30

I put this into the formula for the sum of all of the terms: Sn = n/2 ( a + last term )
S30 = 30/2 ( 12 + 99 )
S30 = 15 x 111
S30 = 1665 which is correct

for part b) I thought all of the terms in the series were 2 digits. If the first term is 12 and the last term is 99, isn't the sum going to be the same as in part a)

I'm confused- any help would be really appreciated! :)
 
What is the smallest two-digit number? Is it 12?

Is the common difference between successive two-digit numbers 3?

Don't be misled to think that (b) continues from (a); but you can use the same kind of thinking. Part (c) can use the results of (a) and (b), but not only those.
 
so the sum is smaller or the same as -10 and bigger or the same as 10?

Is the common difference 1?
 
Yes, the common difference for successive integers is 1. But two-digit numbers are positive; and it's not the sum that is at least 10. Be sure to say what you mean.
 
if all two-digit number are positive, would the formula for the nth term be: nth term = 10 + (n-1)d = 10 + (n-1)1 = 9 + n
 
there are 90 positive two-digit numbers so:

S90 = 90/2 ( 2a + (n-1) d )
S90 = 45 ( 20 + 89 )
S90 = 45 x 109 = 4905 this is the correct answer

Have I used a correct method? :)
 
bumblebee123 said:
there are 90 positive two-digit numbers so:
S90 = 90/2 ( 2a + (n-1) d )
S90 = 45 ( 20 + 89 )
S90 = 45 x 109 = 4905 this is the correct answer
Have I used a correct method? :)
Click to expand...
Keep it simple: sum(1 to 99) - sum(1 to 9)

Btw, both those numbers are also triangular numbers.
 
If you include both 2 digit negative numbers and 2 digit positive numbers what do you think the sum will be?

Only using positive 2 digit numbers you have 10, 11, 12, ..., 99. In this sequence, yes a1=10 and d=1. So the term 99 is what term number?
 
Denis is correct (finally!). We know that there are n numbers from 1 to n. So there are 99 numbers from 1 to 99. But we do not want 1-9 included (we want 10, 11, ... 99) which is 9 numbers. So from 10-99 there are 99-9 = 90 numbers. So what term number is 99?

Personally I would do as Gauss would do! Do you know his trick?
 
Occasionally you might find the term "two-digit numbers" used with both positive and negative in mind, but usually the context is such that "number" means "natural number".

Jomo nicely showed the way to deal with the slight ambiguity: consider what would happen if you take the alternative interpretation, and then realize that that would be an uninteresting problem.
 
okay, that makes much more sense. 99 must be the 108th term ( nth term = 9 + n = 9 + 99 = 108 )?

also, I have no idea who Gauss is! ( I'm only doing IGCSE maths )

It also makes more sense that I wouldn't use negative two-digit numbers and positive two-digit numbers- so thanks!
 
bumblebee123 said:
okay, that makes much more sense. 99 must be the 108th term ( nth term = 9 + n = 9 + 99 = 108 )?

also, I have no idea who Gauss is! ( I'm only doing IGCSE maths )

It also makes more sense that I wouldn't use negative two-digit numbers and positive two-digit numbers- so thanks!
Click to expand...
Seriously, please tell us the sum if you used both positive and negative numbers, please.

Do you really believe that there are 108 numbers from 10 to 99? How many numbers are there from 1 to 100? Which set has more numbers --{1,2,3,..., 100} or {10, 11, 12, ..., 99}? Think about everything you say. Make sure that you truly believe what you say is true. Think about it in different ways to confirm that it makes sense.
 
there are 90 numbers from 10 to 99. So 99 must be the 90th term. I realise now that what I said was stupid haha.
 
for part c) ( hence find the sum of all 2-digit numbers which are not multiples of 3 )

I now know that the sum of all 2-digit numbers = 4905

so, now I need to find out the equation for all 2-digit numbers which are multiples of 3, then subtract it from 4905.

am I on the right track? I'm not sure what to do next.
 
bumblebee123 said:
okay, that makes much more sense. 99 must be the 108th term ( nth term = 9 + n = 9 + 99 = 108 )?

also, I have no idea who Gauss is! ( I'm only doing IGCSE maths )

It also makes more sense that I wouldn't use negative two-digit numbers and positive two-digit numbers- so thanks!
Click to expand...
Dr Gauss would do the following:
S= 10 + 11 + 12 + 13 + ... + 97 + 98 + 99
S= 99 + 98 + 97 + 96 + ... + 12 + 11 + 10 Then Dr Gauss would add the the lines
2S=109 +109 + 109 + 109 + ... + 109 + 109 + 109 90 times
2S= 109*90
S= 109*45 = 4905

And Dr Gauss did something very similar in grade school (1st - 6th grade).
 
bumblebee123 said:
for part c) ( hence find the sum of all 2-digit numbers which are not multiples of 3 )

I now know that the sum of all 2-digit numbers = 4905

so, now I need to find out the equation for all 2-digit numbers which are multiples of 3, then subtract it from 4905.

am I on the right track? I'm not sure what to do next.
Click to expand...
I would 1st list the sequence you want then try to figure out how to sum them. The sequence is 12, 15, 18, ... , 99
Now that you know the sequence get to work!
 
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