如何在序列中找到一个术语?
算术是数学的一部分,包括对数字的研究和对这些数字执行的操作。此类运算是加法、减法、乘法、除法、求幂和求根。该序列是遵循明确模式的对象/数字的集合/枚举集合。与 Set 类似,它可以包含无限(可能)数量的元素(序列的成员称为元素),并且它还允许数字的重复。例如:1、2、3、4、5、6。一个序列中元素的总数称为序列的长度。在上面的示例中,序列的长度为 6。如果存在序列,例如自然数序列 (1,2,3,4,5,6...),则该序列的长度将变为无限。
算术序列
算术序列也是具有确定模式的序列。如果您在序列中取任何数字,然后将其减去前一个数字,结果始终相同或恒定,则它是一个等差数列。在 AP 中,公差是序列中所有连续或连续数字对的恒定差。它由 d 表示。例如:2,4,6,8。在这个序列中,公差是d = 4-2,等于2。同理,6-4 = 2, 8-6 = 2。所以这个序列的公差是2 [公差= a 2 – a 1 ]。
一般来说,存在两种类型的序列,1.增加序列 2.减少序列
- 增加序列:- 如果序列中连续项之间的共同差 (d) 为正,我们可以说该序列是递增的。例如5、10、15、20、25。序列的公差为+5。
- 递减顺序:- 另一方面,如果序列中连续项之间的公差 (d) 为负,则我们可以说序列是递减的。例如,18、14、10、6、2。共同差为-4。
在序列中查找术语
等差数列的第一项记为“a”,然后公差不断相加得到下一项。 AP 的第 n 项是出现在第 n 位的项。在等差数列中,找到第n项的广义公式意味着可以找到任何一项,只需输入该项的值。例如, n=1 将给出1 , n=3 将给出3 ,依此类推。下表给出了一个序列中的术语 a1 = Represents the First term of the sequence a2 = Represents the Second term of the sequence a3 = Represents the Third term of the sequence .. .. an-1 = Represents Second last or (n-1)th term of the sequence an = Represents Last or (n)th term of the sequence an+1 = Represents (n+1)th term of the sequence
为了在序列中找到一个术语,有两种方法可以确定,
- 用于确定序列中的项(例如第 r项)的递归方式/公式
ar = ar-1 + d for r>=2
(for last (nth) term) ⇢ an = an-1 + d for n>=2
这里,a r是序列中的第 r项。 a r-1是序列中的第 (r-1)项,d 是一个共同的差异。
问题:求下一项 5, 8, 11, 14, 17, 20, ?
解决方案:
In this question first, we have to find common difference. common difference = a2 – a1 = 8-5 = 3.
Now using formula, ar = ar-1 + d , here ar-1 =20
So, ar = 20 + 3 = 23.
- 用于确定序列中的项(例如第 r个项)的显式方式/公式
ar = a1 + (r-1)d
(for last (nth) term) ⇢ an = a1 +(n-1)d
这里,a r是序列中的第 r项。 a 1是序列中的第一项,d 是公共差异,r 是第 r个元素的索引。
示例问题
问题 1:在 AP 10, 20, 30, 40, 50, 中找到下一个术语?
解决方案:
Common difference = a2 – a1 = 20-10 = 10
Here a1 = 10 and r or n = 6
Now using formula ar = a1 + (r-1)d = 10 + (6-1)×10 = 60.
问题 2:找到以下序列中的下一项,-7, -1, 5, 11, 17,?
解决方案:
Common difference(d) = a2 – a1 = -1-(-7) = -1+7 = 6.
here a1 = -7 and r or n = 6, here r is the rth element’s index,
Note: n is the last index of the sequence. here index is last so we can say it n as well as r.
Then by using formula , ar = a1 + (r-1)d = -7 + (6-1)×6 = -7 + 30 = 23.
问题 3:找出下列序列 6、11、? 中间的项, 21, 26。
解决方案:
Common difference(d) = a2 – a1 = 11-6 = 5 .
Here, a1 = 6, and r = 3
Then by using formula , ar = a1 + (r-1)d = 6 + (3-1)×5 = 6 + 10 = 16.
问题 4:找出以下序列中的第一项,?, -5, -1, 3, 7。
解决方案:
Common difference(d) = a3 – a2 = -1 – (-5) = -1 +5 = 4
Here , we know the last term in the given an = 7. and a1 is been to be determine. and n = 5 (last term)
Then by using formula , an= a1 + (n-1)×d
7 = a1 + (5-1)*4
7 = a1 + 16
a1 = 7-16 = -9
问题 5:按下列顺序找出下一项。 -1/2,-5/6,-7/6,?
解决方案:
Common difference(d) = a2 – a1 = -5/6 – (-1/2) = -1/3
Here, a1 = -1/2 , r or n = 4
Then by using formula, ar = a1 + (r-1)d = -1/2 + (4-1)*-1/3 = -3/2
问题 6: 找出具有以下两个给定项的数列的公式,a 5 = 20 和 a 20 = 80。
解决方案:
For a5 = 20
Using formula, an = a1+(n-1)*d
so ,for a5 = a1 + (5-1)×d
a5 = a1 + 4d
20 = a1 +4d ⇢ equation(i)
similarly, for a20 = a1 + (20-1)×d
a20 = a1 + 19d
80 = a1 + 19d ⇢ equation(ii)
Solve both equation by Elimination method. multiply equation(i) by -1 and add it to equation(ii) to eliminate a1 ,
-20 = -a1 – 4d (when equation(i) is multiply by -1)
80 = a1 + 19d
On adding both
60 =15d
d = 4
Putting value of d in equation(i)
20 = a1 + 4×4
20 = a1 + 16
a1 = 4
so a1 = 4 and d = 4, the formula we’re looking for is,
an = a1 + (n-1)d
an = 4 + (n-1)×4
an = 4 + 4n -4
an = 4d.