序列是对象的一个枚举集合,在其中允许重复,而顺序事项则形成了一种模式,通过这种模式我们可以识别整个序列。我们可以概括整个序列,称为序列。
例1。偶数序列具有差异4。
解决方案:
2, 6, 10, 14,……………..an.
Here in the above example the first term of sequence is a1=2. And last term is an.
示例2。数字的排列,例如1、1、2、3、5、8,……没有可见的模式,但是序列由给出的递归关系生成。
解决方案:
a1=1, a2=1, a3=2
a3=a1+a2.
an=an-2+an-1 Where n>2
This is called Fibonacci sequence.
例3.一个数字排列,例如2、8、14、20,…具有可见的图案,其顺序由该关系生成。
解决方案:
a1=2, a2=6, a3=14. Common difference between element is constant and is equal to d =a2-a1 or take any two adjacent numbers .
So sequence becomes an=a1+(n-1)*d.
系列:一系列可以高度概括为序列中的所有条款的总和。让一个序列被给定为A 1,A 2,A 3,。 。 。 。 。 。 , n。然后将该表达式称为与给定序列关联的级数。有限级或无级级的依赖关系取决于序列的性质是有限级还是无限级。级数用∑(sigma)表示。因此,系列a1 + a2 + a3 + an = ∑ n k = 1 a k。
算术序列
在算术序列中,一项与下一项之间的绝对差是恒定的。
解释:
序列a 1 ,a 2 ,…a n。被称为算术序列或算术级数如果N + 1 -一个N = d其中d是常数。这是共同的区别。
让我们用第一项A和共同差D进行算术级数。
{A,A + D,A + 2D,A + 3D ……。}
AP的第n个通用项由n = a +(n-1)* d给出。
其中,a是AP的第一项,d是公共差,n是项数。
显式公式
Writing explicit formulas
Let’s take a sequence 6, 16, 26, 36 …76.
The first term of the sequence is 6 and common difference is 10.
We can get any term in the sequence by taking the first term 6 and common difference is 10.
第n个项的计算。
1. | 6 | 6+0.10=6 |
2. | 6+10 | 6+1.10=16 |
3. | 6+10+10 | 6+2.10=20 |
4. | 6+10+10+10 | 6+3.10=36 |
5. | 6+10+10+10+10 | 6+4.10=46 |
6. | 6+10+10+10+10+10 | 6+5.10=56 |
7. | 6+10+10+10+10+10+10 | 6+6.10=66 |
The nth term can be find out easily. The first terms is 6, and we get the difference is 10 in each step.
Above statement can be generalized as 6+(n-1)*d.
In general, this is the standard explicit formula of an arithmetic sequence whose first term is, A, end and common difference is D.
An = A+(n-1)*D.
算术级数的重要属性
- 如果将常数添加到AP的每个项中,则所得序列也是AP
- 如果将一个常数减去AP的每个项,则所得序列也是AP
- 如果AP的每一项都乘以一个常数。然后所得的序列也是AP
- 如果将AP的每个项除以非零常数,则结果序列也是AP
算术级数之和
让我们以a,a + d,a + 2d,a + 3d,….. a +(n-1)* d的顺序给出。
Sn = (n/2)(a + l)
where,
a is the first term
l is the last term of the series and
n is the number of terms in the series
Replacing the last term l by the nth term in equation 3 we get,
nth term = a + (n – 1)d
Sn = (n/2)(a + a + (n – 1)d)
Sn = (n/2)(2a + (n – 1) x d)
例1。按照以下定义的每个顺序写前三个术语:
(i) n = 5n + 2(n-1)
解决方案 :
Solve: (i) put n=1, we get a1=5.1 + 2(1-1) = 5+ 0 =5
Put n=2, we get a2=5.2+2(2-1) =10+2 =12
Put n=3, we get a3=5.3 + 2(3-1) =15 + 4 =19
So first three terms are 5,12, 19.
(ii)A n = 2n +4(n-2)
解决方案 :
Put n=1, we get a1=2.1+4(1-2) =2-4 = -2
Put n=2, we get a2 = 2.2+4(2-2) =4+ 0 =4
Put n=3,we get a3= 2.3 + 4(3-2) =6+4 =10
So the first three terms are -2, 4, 10.
例子2。找到给定表达式的第20个词。
解决方案 :
An=(n-1)(2-n)(3+n)
Solve: put n=20 in given expression,
a20=(20-1)(2-20)(20+3) = 19*-18·23 = -7886.
NOTE: if question asked find nth term of arithmetic sequence then consider the sequence to be an=a+(n-1)*d.
例3:求出所有自然数之和,介于100和1000之间(含100和1000),是5的倍数。
解决方案 :
Solve: first term to be 100 and last terms is 1000 and common difference is 5.
So our formula is Sn=(n/2)[2a+(n-1)*d] .
We need to find the number of terms , so number of terms is given by (1000-100)/5 = 900/5 =180.
S180=(180/2)[2·100 +(180-1)*5].
S180=90*[200+179·5]
S180=90·1095 = 98,550.