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An arithmetic sequence is composed of three numbers, the sum of the three numbers is 30, the sum of the squares of the three numbers is 960, find the three numbers
The sum of these three numbers is 30, and the one in the middle is 10
The first is 10-d and the third is 10 + D
10*（10-d）（10+d）＝960
(10-d）（10+d）=96
d=±2
The two numbers are 8 and 12
If the sum of three numbers is equal to 18 and the sum of their squares is equal to 116, then the three numbers are equal______ ．
Let the sequence be A-D, a, a + D. then we can get A-D + A + A + D = 18, and (A-D) 2 + A2 + (a + D) 2 = 116. The solution is a = 6, d = 2 & nbsp; or & nbsp; a = 6 & nbsp; d = - 2, so the three numbers are 4, 6, 8. So the answer is: 4, 6, 8
The sixth term of the arithmetic sequence (an) is, the sum of the third term and the eighth term is also 5. Find the general term of the sequence (an) and the sum of the first n terms of the sequence (an)
A6 = 5, right
a3+a8=5
Then a6-3d + A6 + 2D = 2 * 5-D = 5
So d = 5
Then A1 = a6-5d = 5-5 * 5 = - 20
an=a1+(n-1)d=-20+5(n-1)=5n-25
So Sn = n (a1 + an) / 2 = n (- 20 + 5n-25) / 2 = 5N (N-9) / 2
In the arithmetic sequence, item 4 plus item 5 plus item 6 equals 4. Item 6 plus item 7 plus item 8 equals 6. What is the tolerance
a1+3d+a1+4d+a1+5d=3a1+12d=4 (1)
a1+5d+a1+6d+a1+7d=3a1 +18d=6 (2)
(2)-(1) 6d=2 d=1/3
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