-
Anonymous users2024-02-11s9=s17, giving a8+a9+a10+.a17=0
So a8+a17=0 a9+a16=0 a10=a15=0 a11+a14=0 a13+a14=0
and s13 is the maximum, so d<0 can be concluded that a14 is negative.
-
Anonymous users2024-02-10Three methods. 1 Let the first term of the difference column be a1 and the tolerance is d (d<0), which is obtained by s9=s17 using the first n terms of the difference column and the formula a1=-25 2d, so sn=na1+d 2*n(n-1)=d 2*[(n-13) 2-169].
Because d<0 so when n=13, s13 is maximum.
2 because s9=s17
So a10+...a17=0
i.e. a10+a17=a11+a16=a12+a15=a13+a14=0
and sn has a maximum, so d<0
i.e. a14 0, a13>0
So the maximum value is s13
3 Because the sum of the first n terms of the equidifference column is a quadratic function with respect to n.
and sn exists at the maximum.
So the sum of the first n terms of the equal difference column is an open downward quadratic function with respect to n.
Because s9=s17
So the axis of symmetry of the quadratic function is 13
i.e. the maximum value is s13
-
Anonymous users2024-02-09If the terms are the equal difference series of Yankai positive number Pei Zao Pei, the tolerance , then ( ) with only a b c d b method one (assignment method): take ; Method 2 (Sorting Inequality) Because each item is a positive number, you may wish to let , then, be known by the sorting inequality; Method 3 (Difference Method): So choose B
-
Anonymous users2024-02-08The first item A1 and the tolerance d must be different, and the dividing point of the positive and negative terms is 1-(A1 D), for example, A1=, D=-2, then 1-(A1 D)=, indicating that the 5th item begins to change signs. ，…
For example, if a1=-20 and d=3, then 1-(a1 d)=, indicating that item 8 begins to change.
-
Anonymous users2024-02-07First of all, the difference series sn=na1+n(n-1)d 2 has 4 quantities sn,n,a1,d denote.
And sn=1, the number of terms n is fixed, the initial value a1 is fixed, and d can be found if 3 quantities are known, that is, the sequence is fixed.
d=2(sn-na1) n(n-1), to ensure that each term of the series is greater than or equal to 0, then a1>=0, d>=0
If the sequence is finite, n(n-1)>=0 is constant, so sn-na1>=0, i.e., 1-na1>=0 so a1>=1 n
sn-s(n-1)=2sn 2 2sn-12sn 2-sn-2sns(n-1)+s(n-1)=2sn 2sn-2sns(n-1)+s(n-1)=0 divided by sns(n-1). >>>More
Three numbers are equal differences, then.
a1+a3=2a2 >>>More
Since it is an equal difference series, so a8-a4=4d, d is the tolerance, then d=-4, from a4=a1+3d, we can know a1=a4-3d=24, from sn=na1+n(n-1)d 2 to get sn=-2n 2+26n >>>More
Trust me, that's right.
Method 1: When there are 2n terms in the equal difference series, the sum of the even terms - the sum of the odd terms = nd (i.e. n * tolerance) and: the sum of the even terms + the sum of the odd terms = the sum of the number series (i.e. the sum of the first 2n terms) So: the sum of the series = 2 * the sum of the odd terms + nd >>>More
1) a1+a12=a6+a7a1+a13=a7*2 can be written as the first term and the tolerance form can be used to prove that s12=(a1+a12)*12 2=(a6+a7)*6s13=(a1+a13)*13 2=a7*13, so a1+2d=12 a6+a7<0, that is, 2a1+11d>0 a7>0, that is, a1+6d<0 is represented by d by the formula A1, that is, a1=12-2d is brought into Eq. respectively: 24+7d>0 12+4d<0 can be solved to obtain -24 70a7<0 to know a6>0, a7<0, and |a6|>|a7|Therefore s1a7>a8>a12, so s6+a7>s7>s8>... >>>More