if you are the school director how many days of work is more practical to choose?
Answers
Step-by-step explanation:
Let's find the sum of a pair consisting of 10 different digits.
\cdots\longrightarrow0+1+\cdots+8+9=\dfrac{1}{2}\times10\times9⋯⟶0+1+⋯+8+9=
2
1
×10×9
\cdots\longrightarrow0+1+\cdots+8+9=45.⋯⟶0+1+⋯+8+9=45.
Every ten digits repeat 11 times, so the sum is,
\cdots\longrightarrow45\times11=495.⋯⟶45×11=495.
So, the answer is,
\cdots\longrightarrow\boxed{495.}⋯⟶
495.
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Where a,d,la,d,l are the first term, common difference, last term, the arithmetic terms are,
\cdots\longrightarrow a,a+d,a+2d,\cdots,l-2d,l-d,l⋯⟶a,a+d,a+2d,⋯,l−2d,l−d,l
Let S_{n}S
n
express the sum of nn consecutive numbers,
\begin{gathered}\begin{aligned}S_{n}&=a+(a+d)+(a+2d)+\cdots+(l-2d)+(l-d)+l\\S_{n}&=l+(l-d)+(l-2d)+\cdots+(a+2d)+(a+d)+a\end{aligned}\end{gathered}
S
n
S
n
=a+(a+d)+(a+2d)+⋯+(l−2d)+(l−d)+l
=l+(l−d)+(l−2d)+⋯+(a+2d)+(a+d)+a
\cdots\longrightarrow 2S_{n}=n(a+l)⋯⟶2S
n
=n(a+l)
\cdots\longrightarrow\boxed{S_{n}=\dfrac{1}{2}n(a+l).}⋯⟶
S
n
=
2
1
n(a+l).
Where a,n,la,n,l are the first term, number of terms, last term, the arithmetic series is,
\cdots\longrightarrow S_{n}=\dfrac{1}{2}n(a+l)⋯⟶S
n
=
2
1
n(a+l)
\cdots\longrightarrow S_{n}=\dfrac{1}{2}\{a+a+(n-1)d\}⋯⟶S
n
=
2
1
{a+a+(n−1)d}
\cdots\longrightarrow\boxed{S_{n}=\dfrac{1}{2}\{2a+(n-1)d\}.}⋯⟶
S
n
=
2
1
{2a+(n−1)d}.