Is it an ARITHMETIC SEQUENCE OR A PROGRESSIVE SEQUENCE

Arithmetic and Geometric progressions.

Is it an ARITHMETIC SEQUENCE OR A PROGRESSIVE SEQUENCE

Postby Learningforcomp » Tue Nov 03, 2020 10:49 am

Hi, I'm practicing arithmetic and progressive sequence when I encountered this problem saying
nnn.png
nnn.png (2.43 KiB) Viewed 1880 times
, at first I thought that it was an arithmetic sequence but the common difference does not quite make sense.
What do I do?
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Re: Is it an ARITHMETIC SEQUENCE OR A PROGRESSIVE SEQUENCE

Postby Guest » Wed Nov 04, 2020 3:32 pm

That is neither an arithmetic progression nor geometric.
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Re: Is it an ARITHMETIC SEQUENCE OR A PROGRESSIVE SEQUENCE

Postby HallsofIvy » Sat Jan 09, 2021 8:55 am

I know what an "arithmetic sequence" is but what do you mean by a "progressive sequence"?

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Re: Is it an ARITHMETIC SEQUENCE OR A PROGRESSIVE SEQUENCE

Postby HallsofIvy » Fri Jan 15, 2021 6:32 pm

Have you looked at the partial sums? 1+ 11= 12, 12+ 111= 123, 123+ 1111= 1234, 1234+ 11111= 12345,
For n up to 9, the sum is 123456789. After that, you have "carry" complications. For example, for n= 10
1111111111+ 123456789= 234567900.

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Re: Is it an ARITHMETIC SEQUENCE OR A PROGRESSIVE SEQUENCE

Postby Guest » Fri Nov 12, 2021 9:11 am

[tex]1+11+111+...+111...1+111...11+111...111=(1)+(1+10)+(1+10+100)+...+(1+10+100+...+100...0)+(1+10+100+...+100...0+100...00)+(1+10+100+...+100...0+100...00+100...000)=(10^{0})+(10^{0}+10^{1})+(10^{0}+10^{1}+10^{2})+...+(10^{0}+10^{1}+10^{2}+...+10^{n-2})+(10^{0}+10^{1}+10^{2}+...+10^{n-2}+10^{n-1})+(10^{0}+10^{1}+10^{2}+...+10^{n-2}+10^{n-1}+10^{n})=\sum_{i=0}^{0}10^i+\sum_{i=0}^{1 }10^i+\sum_{i=0}^{2 }10^i+...+\sum_{i=0}^{n-2}10^i+\sum_{i=0}^{n-1}10^i+\sum_{i=0}^{n}10^i =\frac{1-10^{1}}{1-10}+\frac{1-10^{2}}{1-10}+\frac{1-10^{3}}{1-10}+...+\frac{1-10^{n-1}}{1-10}+\frac{1-10^{n}}{1-10}+\frac{1-10^{n+1}}{1-10}=\frac{1}{1-10}\cdot(1-10^{1}+1-10^{2}+1-10^{3}+...+1-10^{n-1}+1-10^{n}+1-10^{n+1})=\frac{1}{1-10}\cdot((1+1+1+...+1+1+1)-(10^{1}+10^{2}+10^{3}+...+10^{n-1}+10^{n}+10^{n+1}))=\frac{1}{1-10}\cdot((n+1)-10\cdot(10^{0}+10^{1}+10^{2}+...+10^{n-2}+10^{n-1}+10^{n}))=\frac{1}{1-10}\cdot((n+1) \cdot\frac{1-10}{1-10} -10\cdot\sum_{i=0}^{n}10^i)=\frac{1}{1-10}\cdot(\frac{(n+1)\cdot(1-10)}{1-10}-10\cdot\frac{1-10^{n+1}}{1-10})=\frac{1}{1-10}\cdot(\frac{(n+1)\cdot(1-10)}{1-10}-\frac{10\cdot(1-10^{n+1})}{1-10})=\frac{1}{1-10}\cdot(\frac{n+1-10n-10}{1-10}-\frac{10-10^{n+2}}{1-10})=\frac{1}{1-10}\cdot\frac{n+1-10n-10-10+10^{n+2}}{1-10}=\frac{1\cdot( n+1-10n-10-10+10^{n+2})}{(1-10)\cdot(1-10)}=\frac{n+1-10n-10-10+10^{n+2}}{1\cdot(1-10)-10\cdot(1-10)}=\frac{10^{n+2}+n-10n+1-10-10}{1-10-10+100}=\frac{10^{n+2}-9n-19}{81}[/tex]
[tex]n=0[/tex]
[tex]1=\frac{10^{0+2}-9\cdot0-19}{81}=\frac{10^{2}-0-19}{81}=\frac{100-19}{81}=\frac{81}{81}=1[/tex]
[tex]n=1[/tex]
[tex]1+11=\frac{10^{1+2}-9\cdot1-19}{81}=\frac{10^{3}-9-19}{81}=\frac{1000-28}{81}=\frac{972}{81}=12[/tex]
[tex]n=2[/tex]
[tex]1+11+111=\frac{10^{2+2}-9\cdot2-19}{81}=\frac{10^{4}-18-19}{81}=\frac{10000-37}{81}=\frac{9963}{81}=123[/tex]
[tex]...[/tex]
[tex]n=10[/tex]
[tex]1+11+111+...+11111111111=\frac{10^{10+2}-9\cdot10 -19}{81}=\frac{10^{12}-90-19}{81}=\frac{1000000000000-109}{81}=\frac{999999999891}{81}=12345679011[/tex]
[tex]n=11[/tex]
[tex]1+11+111+...+11111111111+111111111111=\frac{10^{11+2}-9\cdot11-19}{81}=\frac{10^{13}-99-19}{81}=\frac{10000000000000-118}{81}=\frac{9999999999882}{81}=123456790122[/tex]
[tex]n=12[/tex]
[tex]1+11+111+...+11111111111+111111111111+1111111111111=\frac{10^{12+2}-9\cdot12-19}{81}=\frac{10^{14}-108-19}{81}=\frac{100000000000000-127}{81}=\frac{99999999999873}{81}=1234567901233[/tex]
[tex]...[/tex]
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Re: Is it an ARITHMETIC SEQUENCE OR A PROGRESSIVE SEQUENCE

Postby Guest » Sat Feb 03, 2024 8:59 pm

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