The first and last term of an AP are 8 and 65 respectively. If the sum of all its terms in 730, find its common difference.
Answer
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Hint: The product of the members of a finite arithmetic progression with an initial element $a_{1},$ common differences d, and $n$ elements in total is determined in a closed expression
This is a generalization from the fact that the product of the progression $1 \times 2 \times \cdots \times n$ is given by the factorial n ! and
that the product $m \times(m+1) \times(m+2) \times \cdots \times(n-2) \times(n-1) \times n$ for positive integers $m$ and $n$ is given by$\dfrac{n !}{(m-1) !}$
Complete step-by-step answer:
In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence
of numbers such that the difference between the consecutive terms is constant. For instance,
the sequence 5,7,9,11,13, 15,…is an arithmetic progression with a common difference of 2 .
If the initial term of an arithmetic progression is $a_{1}$ and the common difference of successive
members are d, then the nth term of the sequence $\left(a_{n}\right)$ is given by:
$a_{n}=a_{1}+(n-1) d$ and in general,$a_{n}=a_{m}+(n-m) d$.
Given $730=\dfrac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{l})$
$\mathrm{l}=65$
$\mathrm{a}=8$
$\Rightarrow$ $730=\dfrac{\mathrm{n}}{2}(8+65)$
$\Rightarrow$ $730=\dfrac{\mathrm{n}}{2}(73)$
$\Rightarrow$ $10=\dfrac{\mathrm{n}}{2}$
$20=\mathrm{n}$
Now $65=\mathrm{l}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$
$\Rightarrow$ $65=8+(20-1) \mathrm{d}$
$\Rightarrow$ $65-8=19 \mathrm{d}$
$\Rightarrow$ $57=19 \mathrm{d}$
$\Rightarrow$ $\mathrm{d}=\dfrac{57}{19}=3$
The common difference is 3.
Note: A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series. The standard deviation of any arithmetic progression can be
calculated as $\sigma=|d| \sqrt{\dfrac{(n-1)(n+1)}{12}}$
where $n$ is the number of terms in the progression and $d$ is the common difference between terms. The formula is very similar to the standard deviation of a discrete uniform distribution.
This is a generalization from the fact that the product of the progression $1 \times 2 \times \cdots \times n$ is given by the factorial n ! and
that the product $m \times(m+1) \times(m+2) \times \cdots \times(n-2) \times(n-1) \times n$ for positive integers $m$ and $n$ is given by$\dfrac{n !}{(m-1) !}$
Complete step-by-step answer:
In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence
of numbers such that the difference between the consecutive terms is constant. For instance,
the sequence 5,7,9,11,13, 15,…is an arithmetic progression with a common difference of 2 .
If the initial term of an arithmetic progression is $a_{1}$ and the common difference of successive
members are d, then the nth term of the sequence $\left(a_{n}\right)$ is given by:
$a_{n}=a_{1}+(n-1) d$ and in general,$a_{n}=a_{m}+(n-m) d$.
Given $730=\dfrac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{l})$
$\mathrm{l}=65$
$\mathrm{a}=8$
$\Rightarrow$ $730=\dfrac{\mathrm{n}}{2}(8+65)$
$\Rightarrow$ $730=\dfrac{\mathrm{n}}{2}(73)$
$\Rightarrow$ $10=\dfrac{\mathrm{n}}{2}$
$20=\mathrm{n}$
Now $65=\mathrm{l}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$
$\Rightarrow$ $65=8+(20-1) \mathrm{d}$
$\Rightarrow$ $65-8=19 \mathrm{d}$
$\Rightarrow$ $57=19 \mathrm{d}$
$\Rightarrow$ $\mathrm{d}=\dfrac{57}{19}=3$
The common difference is 3.
Note: A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series. The standard deviation of any arithmetic progression can be
calculated as $\sigma=|d| \sqrt{\dfrac{(n-1)(n+1)}{12}}$
where $n$ is the number of terms in the progression and $d$ is the common difference between terms. The formula is very similar to the standard deviation of a discrete uniform distribution.
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