a 1+a2=2a 1+d=4
a3+a4=2a 1+5d= 12
4d=8，d=2，a 1= 1
a5+a6 = 2a 1+9d = 2+ 18 = 20
What is the sum of integers between 2.50 and 350 with the last digit of 1?
(5 1+34 1)*30= 1028 1
3. In a positive geometric progression, if a4a5=6, a1AAA7A8 equals _ _?
a 1a2a7a8=(a4a5)^2=36
4. Given the numbers 2, 5 and 8 in the series, what is the number of the series with twice the number 5?
2 root number 5= root number 20
20=2+(n- 1)3=3n- 1
n=7
5. If A, B and C are geometric series, what is the number of alternating current between the image of function y=ax +bx+c and the X axis?
A, b and c do geometric series.
So b 2 = AC
b^2-4ac=b^2-4b^2=-3b^2<； 0
So the intersection is 0.
6. In arithmetic progression, if the sum of the first five items is S5=20, then a3=___?
S5=5a 1+ 10d=20，a 1+2d=4
a3=a 1+2d=4
7. The first n terms of geometric series are marked as Sn. If S4 = 2 and S8 = 6, then S 12 = _ _?
s4/s8=( 1-q^4)/( 1-q^8)= 1/( 1-q^4)= 1/3
q^4=2
q^ 12=8
s4/s 12=( 1-q^4)/( 1-q^ 12)=( 1-2)/( 1-8)= 1/7
2/S 12= 1/7
S 12= 14
8. If the series is arithmetic progression, the first item is a1>; 0，a 2007+a 2008 & gt； 0, a2007 a2008 & lt0, then what is the largest natural number that makes the first n terms and Sn > 0 hold?
a2007 a2008 & lt0
So in 2008,
a2007 & gt-a 2008 & gt； 0
So s 4015 >; 0
9. It is known that in geometric series, a3=3, a 10=384, then the general term an=____?
a 1+2d=3
a 1+9d=384
7d=38 1
d=38 1/7
a 1=-74 1/7
an =-74 1/7+(n- 1)38 1/7
10. in arithmetic progression, a 1 and a2 are two roots of the equation X -3x-5=0, then A5+A8 = _ _ _?
x^2-3x-5=0
a 1，2=(3 √29)/2
When a 1=(3+√29)/2, d=√29, a5+a8=3+ 12√29.
When a 1=(3-√29)/2, d=-√29, a5+a8=3- 12√29.
1 1. In arithmetic progression, A 1+A2+A3 =-24, A 18+A 19+A20 = 78, so the first 20 items and S20 = _ _?
a2=a 1+d=-8
a 19 = a 1+ 18d = 26
d=2，a 1=- 10
S20=-200+ 190*2= 180
12. The first n items of the sequence are recorded as Sn. If Sn=n squared -3n+ 1, then an = _ _ _?
a 1=- 1
an=n^2-3n+ 1-(n- 1)^2+3(n- 1)- 1=2n-4
13. in arithmetic progression, the tolerance d is not equal to 0, and A 1, A3 and A9 are in geometric series, so (a3+a6+a9)/(a4+a7+a10) = _ _?
(a 1+2d)^2=a 1(a 1+8d)
d=a 1
(a3+a6+a9)/(a4+a7+a 10)
=(3a 1+ 15d)/(3a 1+ 18d)
= 18d/2 1d
= 18/2 1
fast
Tired
Give it to me! ! !