If the sum of odd and even terms of an arithmetic sequence with even terms is 24 and 30 respectively, and the last term of the sequence is 10 more than the first term, how many terms are there in the sequence?

Suppose that the sequence has n terms and the tolerance is D, then
an-a1=10 ①
an=a1+(n-1)d ②
Because the sum of odd and even terms is 24 and 30 respectively
So we have dn / 2 = 30-24
dn=12
Substituting (2) into (1)
ND-D = 10
Then 12-d = 10
d=2
So n = 6
There are six items in this sequence

Sequence 1,1,2,3,5,8,13,21, ·, is the 500th number odd or even?

The first two numbers are odd, so the third is even; so the fourth is odd; the fifth is odd, the sixth is even, the seventh is odd, the eighth is odd, the ninth is even; the tenth is odd; the eleventh is even; the twelfth is odd From the first law of odd, odd and even cycles, the 500th law is odd 500 divided by 3 and remaining 2

Finding the difference between the sum of all odd numbers and the sum of all even numbers in natural numbers from 1 to 1000

Solution 1: (1 + 999) * 500 / 2 = 250000 (odd sum) (2 + 1000) * 500 / 2 = 250500 (even sum) 250000-250500 = - 500 (difference) application (first term + last term) * number of terms / 2 = solution 2: 1 + 3 + 5 + +999-（2+4+6+…… +1000）＝（1-2）+（3-4)+…… +（999-10...

1,1,2,3,5,8,13 is the 37th number odd or even, the 126th number?

If every three numbers in a sequence are divided into a group, then the third number is even,
So low 37 is odd and 126 is even

At 2, 4, 6, 8, 10 In the series of even numbers, the number four appears several times
Come on

100
There are 50 in 4 positions;
5 * 5 = 25 in 4 out of 10;
There are 4 in 100 places, 5 * 5 = 25;
100 in total

Arrange the even number sequence starting from 2 into a triangle number table as shown in the figure below, then what row and number is the even number 80 in the table?
two
4 6
8 10 12
14 16 18 20
...........

First of all, if there is a number in the first row, 2 in the second row, 3 in the third row, and there are n rows, then there are always n (n + 1)] / 2 numbers in the n row (this is very simple). If n = 12, there are 78 numbers in total, n = 8, 36 numbers in total, and N = 9, there are 45 numbers in total

Use the six numbers 0,1,2,3,4,5 to form a four digit even number without repetition, and arrange them from small to large in order to form a sequence, then the 71st of the sequence
A.3140 B.3254 C.3012 D.3410

If the thousand digit is 1, the hundred digit is 0, and the individual digit is 2 or 4, then the ten digit can be the remaining three choices after the individual digit is determined, so there are a total of 2 × 3 = 6
If the thousand digit is 1, the hundred digit is 2, and the individual digit is 0 or 4, then the ten digit can be the remaining three choices after the individual digit is determined, so there are a total of 2 × 3 = 6
If the thousand digit is 1, the hundred digit is 3, and the individual digit is 0 or 2 or 4, then the ten digit can be the remaining three choices after the individual digit is determined, so there are a total of 3 × 3 = 9
If the thousand digit is 1, the hundred digit is 4, and the individual digit is 2 or 0, then the ten digit can be the remaining three choices after the individual digit is determined, so there are a total of 2 × 3 = 6
If the thousand digit is 1, the hundred digit is 5, and the individual digit is 0 or 2 or 4, then the ten digit can be the remaining three choices after the individual digit is determined, so there are a total of 3 × 3 = 9
When 1000 is 1, there are 36 kinds
The thousand is 2, the hundred is 0, and the individual can only be 4. Then there are three choices for the ten, so there are 1 × 3 = 3
If the thousand digit is 2, the hundred digit is 1, and the individual digit is 0 or 4, then the ten digit can be the remaining three choices after the individual digit is determined, so there are a total of 2 × 3 = 6
If the thousand digit is 2, the hundred digit is 3, and the individual digit is 0 or 4, then the ten digit can be the remaining three choices after the individual digit is determined, so there are a total of 2 × 3 = 6
If the thousand bits are 2, the hundred bits are 4, and the individual bits can only be 0, then the ten bits can be the remaining three choices, so there are 1 × 3 = 3
If the thousand digit is 2, the hundred digit is 5, and the individual digit is 0 or 4, then the ten digit can be the remaining three choices after the individual digit is determined, so there are a total of 2 × 3 = 6
When the number of thousand is 2, there are 24 kinds
36 + 24 = 60
If the thousand digit is 3, the hundred digit is 0, and the individual digit is 2 or 4, then the ten digit can be the remaining three choices after the individual digit is determined, so there are a total of 2 × 3 = 6
If the thousand digit is 3, the hundred digit is 1, and the individual digit is 0 or 2 or 4, then the ten digit can be the remaining three choices after the individual digit is determined, so there are a total of 3 × 3 = 9
60+6+9=75>71
So the 71st must be in 31 () ()
Write these down: 3102 3104 3120 3124 3140
So the answer is a
How hard it is!

1. Three continuous even numbers, the middle one is n, then the sum of these three numbers is 2. Observe the following sequence: 1,2,4,8,16,32, etc., and conclude that the nth term is

1. If the middle is n, then the front one is N-X and the back one is n + X. The sum of the three numbers is n + N-X + N + x = 3N
2、1,2,4,8,16,32,
The rule is 0 th power of 2, 1 th power of 2, 2 th power of 2, 3 th power of 2, 4 th power of 2
2^(n-1)

Sequence: 3 6 18 27
Excuse me: = how much

=11
3+3=6
6+5=11
11+7=18
18+9=27

What are the 5th, 6th and 7th of sequence 1, 3, 8, 18 and 38?
The second is 2 more than the first, the third is 5 more than the second, and the fourth is 10 more than the third

1*2+1=3
3*2+2=8
8*2+2=18
18*2+2=38
I think the next number should be 38 * 2 + 2 = 78, right?

If the sum of odd and even terms of an arithmetic sequence with even terms is 24 and 30 respectively, and the last term of the sequence is 10 more than the first term, how many terms are there in the sequence?