Sequence (31 13:19:16) The first term X1 = 3, the general term xn = 2np + NQ (n belongs to n *, PQ is a constant) of the sequence {xn} is known, and x1, x4, X5 form an arithmetic sequence (1) The values of P, q are as follows (2) The formula of the first n terms of sequence {xn} and Sn  

Sequence (31 13:19:16) The first term X1 = 3, the general term xn = 2np + NQ (n belongs to n *, PQ is a constant) of the sequence {xn} is known, and x1, x4, X5 form an arithmetic sequence (1) The values of P, q are as follows (2) The formula of the first n terms of sequence {xn} and Sn  

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What are the greatest common factor and the least common multiple of 33 and 22

Greatest common factor 11 least common multiple 66

What are the greatest common factor and the least common multiple of 22 and 23

The greatest common factor of 22 and 23 is 1;
The least common multiple of 22 and 23 is 253

The 10th item of arithmetic series 20,17,14

It can be concluded that the first item is: 20, and the tolerance is - 3
Therefore, it can be concluded that:
a10=a1+(10-1)d
=20+9x(-3)
=-7

The sum of the last 200 terms of the arithmetic sequence 200199 and 2 / 3,..., - 100 is equal to?

d=-1/3
Reverse the sequence
Then the tolerance is the opposite
So A1 = - 100
d=1/3
Then A200 = a1 + 199d = - 101 / 3
So s = (a1 + A200) * 200 / 2 = - 40100 / 3

What is the sum of the first 20 terms of the arithmetic sequence 2,6,10?

800

There are two arithmetic sequences 2, 6, 10 , 190 and 2, 8, 14 The common terms of the two arithmetic sequences form a new sequence from small to large, and the sum of each item of the new sequence is calculated

There are two arithmetic sequences 2, 6, 10 , 190 and 2, 8, 14 The common terms of these two arithmetic sequences form a new sequence from small to large, 2, 14, 26, 38, 50 There are 182 − 212 + 1 = 16, which are also arithmetic sequences. The sum of them is 2 + 1822 × 16 = 1472. The sum of the new series is 1472

Given that the third term of arithmetic sequence is - 4, the sixth term is 2, find its tenth term

a6=2 a3=-4
a6-a3=3d d=2
a3=a1+2d
a1=-8
a10=a1+9d=-8+9*2=10

A student measures the length of a new "Zhonghua" drawing pencil. The results of four times of measurement are L1 = 16.58cm, L2 = 16.57cm, L3 = 16.55cm, L4 = 16.56cm. The graduation value of the scale used by the student is___ The measurement result is___ ．

(1) Because when using the scale to read, it needs to estimate one digit when reading the graduation value, so the second digit unit from the right is the graduation value, so the graduation value of the ruler used is 1 mm; (2) taking the average of the four measurement results is the record result: 16.58cm + 16.57cm + 16.55cm + 16.56cm4 ≈ 16.57cm. So the answer is: 1 mm; 16.57cm

How to calculate 56 × (4.5-2.6) + 5.6

56x2-56x0.1+5.6=112-5.6+5.6=112